**CHAPTER **

**6 **

**Normal distribution **

Are the diameters of the rods exactly equal?

## I

### In your earlier studies of statistics, frequency distributions were considered and it was stated

### that, although they are sometimes of importance in themselves, they are mainly important

### in providing information about the population from which the sample is drawn.

**a b ****a b **

Figure 6-1: Histogram Figure 6-2: Frequency curve

### Consider a relative frequency distribution of a continuous variable, represented by the

### histogram (Figure 6-1). It is not difficult to imagine that, if the sample size is increased

### indefinitely, it will approach the population size and, if the class intervals are made as small

### as possible, the histogram will approach a well defined smooth curve (Figure 6-2). Such a

### curve is called afrequency curve or probability density curve and it provides us with

### information about a population in much the same way as the histogram does about the

### sample. The shaded area in the histogram represents the relative frequency with which the

### variable lies between the values a and bin the sample, and corresponds to the shaded area

### in the frequency curve which represents the relative frequency or probability with which the

### variable Hes between the values a and bin the population. If the probability density curve

### is drawn so that the total area under the curve is unity and if the equation of the curve

**y **

**y**

### = f(x) is known, then the probability that X

### lies in the interval a to b can be calculated

### by the process of integration:

**Pr(a < **

**Pr(a <**

**X **

**X**

### < b) =

**[**

**[**

**b **

**b**

**f(x)dx. **

**f(x)dx.**

### The mean and standard deviation of a population are denoted by Greek letters µ and a

### respectively and are called parameters to distinguish them from the mean and standard

### deviation of a sample which are denoted by Roman letters x ands respectively and are called

**statistics. **

**statistics.**

**6.1 The normal distribution **

*y *

**a **

### One of the most important examples of a probability distribution of a continuous variable

### is the

*normal *

### distribution. If a random variable,

*X, *

### is normally distributed, its frequency

### curve has a typical symmetrical bell shape as.shown in Figure 6-3. This curve has been

### found to give an adequate fit to a great variety of frequency distributions. The heights of

### children of a certain age, the diameters of metal cylinders, the number of burning hours

### of electric light globes manufactured by a particular firm, the Intelligence Quotient of

### children in a certain area, are a few examples of distributions which are approximately

### normal.

### The normal distribution is defined by the equation of its frequency curve:

### where:

### and:

### 1

### -! (�)'

*Y*

**= **

*--e *

*i u*

*ufu *

### µ, = the mean value of X in the population

### u

**= **

### the standard deviation of X in the population.

### How the equation of this curve is derived is beyond the scope of this book, but the following

### properties should be noted.

### (i) The curve extends, theor"etically at least, to infinity in either direction and so

*X *

### can

### assume all values.

### (ii) The curve is symmetrical about the ordinate

*x *

**= µ, and so the mean, mode and median**

### of the normal distribution coincide.

**(iii)**

### Practically all of the population (about 99. 7 per cent) lies in the interval µ, ±

*3a; *

### about

### 95 per cent lies in the interval µ,

**± **

*2a; *

### about

### j

### of the population lies in the interval

### µ,

**± **

*(1.*

**(iv)**

### The total area under the curve is unity.

### (v) The probability that

*X *

### lies in the interval

*a *

### to

*b *

### is equal to the area under the curve

### bounded by the ordinates

*x *

### =

*a *

### and

*x *

### =

*b.*

**6.2 Standard normal curve **

### Normal distribution curves will differ in location and degree of spread according to the

### values of the parameters µ, and a respectively.

*y *

*y *

*y *

**X **_{µ } **X **

Figure 6-4 Figure 6-5

### In Figure 6-4, the curves differ in location but have the same degree of spread, i.e.

*a *

### is the

### same for both, but µ, is different.

**It **

### would appear then, that the area for any interval would require special determination for

### each particular curve. However, this is not the case, since all normal curves can be

### transformed into a standard normal curve by putting µ =

**0 **

### and

*a *

### = l. The equation of

### the standard curve is:

1 _! ,'

**Y **

**Y**

**= **

**--e**

**--e**

**2**

*-J2ir *

### To transform the normal equation to the standard normal equation put:

*x-µ*

*z *

### = --and

**Y= ay.**

**Y= ay.**

*(1 *

### Whenz

**= **

### 0,

**Y = -**

**Y = -**

### 1-e

**0**

_{= -}

_{= -}

### 1-

### =

_{0.40 }

*-J2ir *

*-J2ir *

### Whenz

**= ± **

### 1,

**Y **

**Y**

**= **

_{v21r }

**},_e-f **

**},_e-f**

### =

### 0.24

### When

**z **

**z**

**= ± **

**2, **

_{Y }

_{Y }**= . **

### },_

**e **

**e**

### - 2

### =

**0.05**

### v21r

### When

**z **

**z**

**= ± **

### 3,

**Y **

**Y**

**= . **

_{v21r }

### },_

**e **

**e**

### -

4-5### = 0.004 (using a calculator)

### For example, to evaluate -,/2ir

**e **

**e**

### -

4-5,### we can proceed as follows:

**(g) **

**a a e • **

**111:1@ **

### o

**CID 11111111 **

**e **

### -3

**-2**

-1
Figure 6-6: Standard normal curve

*z *

**0.4 **

0.3

**0.2 **

0.1

**z, **

0

### 0°0044

2 3 **z **

### The total area under this curve is unity but the area from - oo to any positive value of z

### is given in the normal probability tables provided (see page 178).

### To find the area under any normal curve between the ordinates x1 and X2, we find the

### corresponding values of z1 and z2 from the transformation formulae:

### X1 - µ

*-*

**X2**### µ

### z1

**= **

### ---andz2

_{a }

_{a }

**= ---**

_{a }

_{a }

**Example 1 **

### The heights of VCE students in Victoria may be considered to be normally distributed with

### a mean of 170 cm and a standard deviation of

*5 *

### cm.

### a What is the probability that a student, selected at random, has a height between 174 cm

### and 178 cm?

### b Out of a group of 150 VCE students, how many would be expected to have a height less

### than 164 cm?

### c What proportion of students would be expected to have heights deviating from the mean

### by more than two standard deviations?

*y *

*A *

155 160 :165 170 175 180 185 **X**

### -3

-2**:-1**0

### '

: 1 2### 3

**z **

**z**

-1'.2 o.'8 1.6

Figure 6-7

### a The shaded area,

**A, **

**A,**

### in Figure 6-7 measures the required probability.

### Whenx = 174:

### Whenx

**= **

### 178:

### z = X - /J, = 174 - 170 = 0.8

_{a }

_{a }

_{5 }

_{5 }

**Pr(X **

**Pr(X**

### <

### 174)

**= **

**Pr(z **

**Pr(z**

### <

### 0.8)

**= **

### 0.7881 (from tables)

### z =

**X �**### /J, = 178 ; 170 = 1.6

**Pr(X **

**Pr(X**

### <

### 178) =

**Pr(z **

**Pr(z**

### <

### 1.6)

### = 0.9452 (from tables)

### Pr(174

*<X *

### <

### 178)

**= **

### 0.9452 - 0.7881

### = 0.1571

### b The shaded area,

**B, **

**B,**

### in Figure 6-7 measures the probability of a student having

### a height less than 164 cm.

### Whenx = 164:

_{4 }

_{0 }

### z = X - µ, = 16 - 17 = _ 1.2

_{a }

_{a }

_{5 }

_{5 }

**Pr(X **

**Pr(X**

### <

### 164)

**= **

**Pr(z **

**Pr(z**

### < -

### 1.2)

### =

**Pr(z **

**Pr(z**

### >

### 1.2) (from symmetry of curve)

### = 1 -

**Pr(z **

**Pr(z**

### <

### 1.2)

### = 1 - 0.8849

**= **

### 0.1151

### Expected number

**= **

### 150 x 0.1151

*y *

**155 ** **160 ** **165 ** **170 ** **175 ** **180 ** **185 X **

**Figure 6-8 ** **-3 ** **-2** **-1** **0 ** **2 ** **3 z **

### c The required proportion is the sum of the areas

*C *

### and Din Figure 6-8. By

### symmetry, these areas are equal. For two standard deviations above or below the

### mean,z

**= ± **

### 2.

### Pr

*(X> µ *

### +

### 2o-) =

*Pr(z> *

### 2)

**= **

### 1 - Pr(z<2)

**= **

### 1 - 0.9772 (from tables)

**= **

### 0.0228

*Pr(X> µ *

### +

*2a or<µ *

### - 2a)

**= **

### 2 x 0.0228 (from symmetry)

### = 0.0456

### This means that about

*5 *

### OJo of the students have heights deviating from the mean

### by more than two standard deviations, or that about 95% of students have

### heights within two standard deviations from the mean. This is one of the

### characteristics of the normal distribution. Verify that about� of the population

### lies within one standard deviation from the mean and that practically all of the

### population (about 99.7%) lies within three standard deviations from the mean.

**Example 2 **

### A lathe turns out brass cylinders with a mean diameter of 2.00 cm and a standard deviation

### of 0.04 cm. Assuming that the distribution of diameters is normal, find the limits to the

### acceptable diameters if, on checking, it is found that five per cent in the long run are rejected

### because they are oversize and five per cent are rejected because they are undersize.

### Each of the shaded areas in Figure 6-9 is five per cent of the total area. This is an

### inverse type of problem in which the proportions are given and the x-values of the

### inside ends of the shaded portions are to be found. Using the inverse normal

### distribution tables on page 178, we find the z value such that the area from -oo to

### this z value is 0.95. Its value is 1.6449. By symmetry, the other z value is -1.6449.

**Figure 6-9 **

**1.92 ** **1.96 **

**-2** **-1**

**-1.6449**

*y *

**2.00 **
**0 **

,

**1.6449 **
**0.05 **_{, }

I / ---...__{_..._ }

### \

_{z -}

*x-*

*µ·"-...*

� 7"-,..- ')

**'Z **

V
**'Z**

"'- .

### + 1 6449

*:·x� *

### 2·00

### - .

_{0.04 }

**X **

**= **

### 2.00 ±

### 1.6449 X 0.04

**= **

### 2.00

**± **

### 0.07

**= **

### 1.93 or 2.07 (cm)

### These are the acceptable limits.

**Example 3 **

### The mean life of a certain type of television tube is 10 000 hours with a standard deviation

### of 1000 hours. Assuming the distribution of lifetimes, X, is normal, find the probability that

### Xis less than any specified value

*x. *

### Using integral multiples of the standard deviation, plot

### the cumulative probability curve.

Figure 6-1 O

*y *

**Pr (X< x) **

### �-+--'�--,...---t--��-i----i---1�x

_{7 }-3

8 -2

9 -1

10 * X * 11

**0**

12 2

13 hours ('000)
3 **z **

### The shaded area of Figure 6-10 shows the probability that Xis less than

*x, *

### where

### the values of the variable lie almost certainly between 7000 and 13 000.

### When

*x *

**= **

### 7000:

**z **

**z**

### = - 3 and Pr(X

### <

### 7000) = 0.0013 (from tables)

### When

*x *

**= **

### 8000:

**z **

**z**

**= -**

### 2 and Pr(X

### <

### 8000)

**= **

### 0.0228

### Similarly, for

*x *

**= **

### 9000, 10 000, .. . 13 000 as shown in the following table:

**x('OOO) ** 7 8 9 10 11 12 13

**Pr(X<x) ** 0.0013 0.0228 0.1587 0.5 0.8413 0.9772 0.9987

### This table gives a cumulative probability distribution of a normal variable, X. It

### is similar to a cumulative relative frequency distribution. From the table we can see,

### for example, that 84.13 per cent of tubes have a lifetime less than 11 000 hours.

### Figure 6-11 shows the cumulative probability curve from which the different

### quantiles and relative frequencies can be calculated approximately. For example,

### about 28 per cent of tubes have a lifetime less than about 9400 hours. The 0.6

### quantile is approximately 10 250.

**1.0 **

### �

**0.8**

**0.6**

**0.4 **

**0.28-0.2 **

**0 **
**7 **

*V *

*----I*

### ---

### ---

### ---

_{i'i }

_{i'i }

*I *

### ---

### ---

### _/

*V *

### '

### ---

### ' '

**8 ** **9 : 10 **

### '

**11**

**12**

**13**hours

**('000)**

Figure 6-11 **9_'4 10.25 **

**Example4 **

### A machine makes metal rods with a mean length of 50 cm and a standard deviation of 1 cm .

### Assume that the distribution of lengths is normal.

### a What proportion of rods whose length is greater than 49 cm will have a length in excess

### of 50cm?

### b If five rods are selected at random, what is the probability that not more than one of

### these rods will have a length greater than 49 cm?

**49 **
**-1 **

*y *

**50 **
**0 **

**51 ** **52 **
**2 **

**53 X **
**3**

*z *

Figure 6-12

### a

**Pr(X> **

**Pr(X>**

### 49)

**= **

**Pr(z **

**Pr(z**

### >

### -1)

**= **

**Pr(z **

**Pr(z**

### <

### 1)

**= **

### 0.8413

**Pr(X> **

**Pr(X>**

### 50)

**= **

### 0.5

### This question involves conditional probability.

### Using the formula:

_{Pr(A }

_{Pr(A }_{n }

_{B) }

_{B) }**Pr(B **

**Pr(B**

### I

**A) **

**A)**

**= **

_{Pr(A)}

### we get:

### d

**Pr(X> **

**Pr(X>**

### 50)

**Pr(X>.5 **

**Pr(X>.5**

### IX> 49) =

_{Pr(X> }

_{Pr(X> }_{49). Why?}

### ---

_{0.5 }

**= **

_{0.8413 }

_{0.8413 }

**= **

### 0.5943

### Or, simply using the geometry of the situation as shown in Figure 6-12:

### NORMAL DISTRIBUTION 175

### b Since each rod has the same probability 0.8413 of having length greater than

### 49 cm, we are dealing with a

*binomial *

### variable,

*Y, *

### withp

**= **

### 0.8413,

**q **

**q**

**=**

### 0.1587

### and

**n **

**n**

**= **

### 5.

**Pr(Y � **

**Pr(Y �**

### 1)

**= **

**Pr(Y **

**Pr(Y**

**= **

### 0)

### +

**Pr(Y **

**Pr(Y**

**= **

### 1)

**Exercises 6a **

**= **

### (0.1587)

5_{+ (\0) (0.1587)}

4_{(0.8413) }

**= **

### 0.0001 + 0.0053

**= **

### 0.0054

### 1 Plot the curve of the normal distribution:

### Y

_{= }

_{= }

_{_l _ }

_{_l _ }

_{crfiir }

### e-½(7)'

**whenµ **

**whenµ**

**= **

### 20 and

**<1 **

**<1**

**= **

### 5 and, by counting squares, verify that the area under the curve

### is approximately one square unit.

*:£ *

### A normal variable has meanµ and standard deviation

**<1. **

**<1.**

### What is the probability that

### any value of the variable, randomly selected, lies betweenµ +

**0.4<1 **

**0.4<1**

### and µ + 2.6<1?

### 3 A manufacturer of electric light globes finds that these articles have an average life of

### 1200 burning hours with a standard deviation of 200 hours. Assuming that the

### distribution of life-times is normal:

### a what is the probability of a globe selected at random having a life between 1240 and

### 1320 hours?

### b out of a batch of 200 globes, how many would be expected to fail in the first 880

### burning hours?

### c what proportion of globes manufactured would be expected to have a life less than

### 1100 hours or more than 1460 hours?

*@ *

### Tests on breaking strengths of two different kinds of fibre, one being silk and the other

### a silk-rayon mixture, yielded the following data:

### Silk: mean 10 kg wt; standard deviation 2.5 kg wt.

### Silk-rayon: mean 15 kg wt; standard deviation 5 kg wt.

### Calculate:

### a the probability that a piece of silk, selected at random, will be at least as strong as

### the mean for the silk-rayon mixture.

### b the probability that a piece of silk-rayon selected at random will be no stronger than

### the mean of the silk.

### 5 A machine makes electrical resistors which have a mean resistance of 50 ohms with a

### standard deviation of 2 ohms.

### a Assuming the distribution to be normal, find the proportion of resistors made which

### have resistance of less than 47 .5 ohms.

### b Calculate the limits

**a **

**a**

### and

**b, **

**b,**

### equally spaced on either side of the mean, so that the

### manufacturer can correctly claim that, in the long run, no more than one resistor

### in 500 lies outside these limits.

### · V The local authorities in a certain city install ! 000 electric lamps in the streets of the city.

### a If the lamps have an average life of 2000 burning hours with a standard deviation

### of 400 hours, and the life of the lamps is normally distributed, what number of lamps

### might be expected to fail in the first 1500 burning hours?

### 7 Steel rods are manufactured to be

*5 *

### cm in diameter, but they are acceptable if they are

### between 4.95 and

*5 *

### .05 cm. The manufacturer finds that, in the long run, about four per

### cent are rejected as oversize and four per cent as undersize.

**If **

### the diameters are

### normally distributed, find the distribution's standard deviation.

**8 **

### Speedometers of cars are not accurate. Suppose that, when the speedometer of a

### randomly chosen car registers 60 km/h, the actual speed of the car is a variable having

### a normal distribution with meanµ

**= **

### 62 ana standard deviation a

**= **

### 2.

### What proportion of cars are exceeding 60 km I h when their speedometers register

### 60 km/h?

**9 **

### The 'threshold', or smallest amount, of a certain poison which is sure to kill a rat, is

### known to vary from rat to rat, following a normal distribution with mean 25 .0 mg and

### standard deviation 2.5 mg.

### a Find the proportion of rats that would be killed by a dose of 27 .0 mg.

**b **

### Plot a graph (using integral multiples of the standard deviation) showing how this

### proportion changes as the dose is increased or decreased.

### c Find the smallest dose that would kill 90 per cent of rats.

**d **

### What changes would you expect in the graph if the poison were diluted by adding

### two parts of inert bait to one part of the original poison?

**10 **

### Butter, marketed in 250 g packages, has a weight which is normally distributed with its

### advertised weight of 250 g as the mean. It may be regarded as appreciably underweight

### if the actual weight is less than 225 g. Find the maximum allowable value of the

### standard deviation if, in the long run, not more than one package in 100 is rejected as

### being underweight.

**11 **

### The mean annual income for a sample of 200 persons selected at random from a certain

### industry was $20 800 with a standard deviation of $4160. Of these, eight earned less

### than $272 per week and 24 earned more than $480 Qer week. Does this sample tend to

### confirm or refute the claim that incomes of the population from which this sample was

### selected were normally distributed? Give reasons for your answer.

**12 **

### A firm producing brass washers to a specified thickness of 0.5 cm has found that the

### thickness varies normally about a mean of 0.5 cm with a standard deviation of

### 0.005 cm. All washers with a thickness between 0.49 cm and 0.51 cm are regarded as

### satisfactory. In a batch of 2000 washers, how many would you expect to be rejected?

**13 **

### The average height of male students at a certain university is 170 cm with variance 25.

### What proportion of these students whose height is greater than 160 cm will have a

### height in excess of 170 cm assuming the heights are normally distributed?

**14 **

**If **

### Xis a normally distributed random variable with mean 10, and the probability that

### Xis greater than 12 is 0.1056, find:

### a the standard deviation of X

**b **

*Pr(X> 8) *

### c

*Pr(X> *

### 8

J*X< *

### 12)

**d **

### the value of

*x *

### for which

*Pr(X *

**> **

*x) *

**= **

### 0.85

### 15 .·Xis a normally distributed random variable and the variance of Xis 4. Given that

*Pr(X> *

### 16) = 0.95, findµ, the mean value of

*X *

### and determine:

### a

*Pr(X *

### <

### 16

*IX<µ), *

**b **

*Pr(X *

### <

*µ *

### IX<

### 20).

### /

### 17 A certain population of plants has a distribution of heights, measured in centimetres,

### which is normal with mean 30 cm and standard deviation 2 cm.

**a **

### Cals;ulate the probability that a randomly selected plant will be less than 27 cm in

### height.

**b **

**If **

### five plants are selected at random, what is the probability that, at most, one is less

### ilimn����n

**18 **

### The average life of a certain type of light globe is 1200 h with a standard deviation of

### 240 h.

### a Assuming that the lengths of life of this type of globe are normally distributed,

### complete the following proportionate frequency distribution.

*Length of life *

<480 <720 <960 < 1200 < 1440 < 1680 < 1920
*Proportion of tubes *

**b **

### Use the table above to draw a cumulative proportion curve and, from the graph, find:

**(i)**

### the proportion of tubes which have a life less than 700 h.

**(ii)**

### the 0.9 quantile, stating what it represents.

**19 **

### The wingspan of birds of a particular species has a normal distribution with mean 50 cm

### and standard deviation 5 cm.

### a Find the probability that a randomly selected bird has a wingspan greater than 60 cm.

### b

**If **

### the wingspan is measured to the nearest centimetre, find the probability that a

### randomly selected bird has a wingspan measured as 50 cm.

### 20 The length of a certain species of fish has a normal distribution with mean 30 cm and

### standard deviation 2.5 cm.

**a **

### Find the probability that a randomly selected fish has a length greater than 36 cm.

**b **

**If **

### the lengths of the fish are measured correct to the nearest centimetre, show that

### the probability of a randomly selected fish having a length which is measured as

### 30 cm is about 0.16.

### c

**If **

### five fish are randomly selected, find the probability that exactly two will have their

### lengths measured as 30 cm.

**21 **

### Suppose that the strengths of mass-produced items are normally distributed with mean

µ

### and standard deviation 0.5. The value of

µ### can be controlled by a machine setting.

**If **

### the strength of an item is less than 5, it is classified as defective. Revenue from sales

### of non-defective items is $20 per item, while revenue from defective items is $2 per item.

### The cost of production of items with mean-µ is $2 per item. Find the expected profit

### per item ifµ = 6.

### 2� A machine makes electrical resistors which have a mean resistance of 50 ohms with a

### ·� · standard deviation of 2 ohms.

**23 **

**a **

### Assuming the distribution of resistances to be normal, find the proportion of

### resistors made which have resistance less than 47 .5 ohm�.

**b **

**If **

### IO resistors are selected randomly, what is the probability that no more than one

### will have a resistance less than 47 .5 ohms?

### A chain is made of five links which are selected at random from a population of links.

### The strengths of the links are assumed to be normally distributed with mean of 500 units

### and a standard deviation of IO units. Find the probability that:

### a a randomly selected link has a strength of at least 490 units

### b a chain has a strength of at least 490 units

**c **

### at least two links in a chain have strength of at least 490 units.

**Area under standard normal curve **

**giving area as function of x, x � 0 **

/( **0 ** **1 ** **2 **

*CD *

**4**

**5**

**6**

**7**

**6**

**9**

**1**

**0.0 _{�1000 'Q'.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 4 }**

**0.1**

**,5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 4**

**0.2 0.5793 o.5832 o.5871 o.591 o 0.594�8t) 0.6026 0.6064 0.6103 0.6141 4 **

**0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 4 **

**0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 4 **

**0.5 0.6915 0.6950 0.6985 0.7019 0. 7054 0. 7088 0. 7123 0. 7157 0. 7190 0. 7224 3 **

**0.6 0.7257 0.7291 0.7324 0.7357 0. 7389 0. 7 422 0. 7 454 0.7486 0.7517 0.7549 3 **

**0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 3 **

**0.8 0.7881 o. 791 o o. 7939 o. 7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 3 **

**0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 3 **

**1.0 0.84� 3 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 2 **

**1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.87 49 0.8770 o.8790 o.881 o o.8830 2 **

**1.2 0.8849 0.8869 0:8888 0.8907 0.8925-l�-8962 0.8980 0.8997 Q.9015 2 **

**1.3 0.9032 0.9049 0.9066 0.9082 0.9099 !f.9fl 5 0.9131 0.9147 0.9162 0.9117 2 **

**1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1 **

**1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1 **

**1.6 0.9452 0.9463 0.947 4 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1 **

**1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1 **

**1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1 **

**1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.97 44 0.9750 0.9756 0.9761 0.9767 1 _{' }**

**2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 o **

**2.1 .0.982) 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 o **

**2.2 0.9861 0.9864 0.9868 0.9811 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 o **

**@ **

**0.9893 0.9896 0.9898 ().99011**

_{0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 o }**2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0. 9932 0. 9934 0. 9936 o**

, .. *)' * )

**2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 o **

**2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 o **

**2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.997 4 o **

**2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 o **

**2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 o.9985 o.9986 o.9986 o **

### :r.o

**0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990 o**

**3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993 o **

**3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 o **

**3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 o **

**3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998 o **

**3.6 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 o **

**3.8 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 o **

**3.9 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 o **

**Inverse normal tables **

**0.50 **

**0.0000**

**0.60 **

**0.2533**

**0.70 **

**0.5244**

**0.80 **

**0.8416**

**0.90 **

**0.51 **

**0.0251,**

**0.61 **

**0.2793**

**0.71 **

**0.5534**

**0.81 **

**0.8779**

**0.91 **

**0.52 **

**0.0502**

**0.62 **

**0.3055**

**0.72 **

**0.5828**

**0.82 **

**0.9154**

**0.92 **

**0.53 **

**0.0753**

**0.63 **

**0.3319**

**0.73 **

**0.6128**

**0.83 **

**0.9542**

**0.93 **

**0.54 **

**0.1004**

**0.64 **

**0.3585**

**0.74 **

**0.6433**

**0.84 **

**0.9945**

**0.94 **

**0.55 **

**0.1257**

**0.65 **

**0.3853**

**0.75 **

**0.6745**

**0.85 **

**1.0364**

**0.95 **

**0.56 **

**0.1510**

**0.66 **

**0.4125**

**0.76 **

**0.7063**

**0.86 **

**1.0803**

**0.96 **

**0.57 **

**0.1764**

**0.67 **

**0.4399**

**0.77 **

**0.7388**

**0.87 **

**1.1264**

**0.97 **

**0.58 **

**0.2019**

**0.68 **

**0.4677**

**0.78 **

**0.7722**

**0.88 **

**1.1750**

**0.975 **

**0.59 **

**0.2275**

**0.69 **

**0.4959**

**0.79 **

**0.8064**

**0.89 **

**1.2265**

**0.98 **

**Mean dlfferen es **

**2 (!: 4 5 6 ** **7 8 9 **

**8 12 76 20 24 28 32 36 **
**8 12 16 20 24 28 32 36 **
**8 12 15 19 23 27 31 35 **
**8 11 15 19 22 26 30 34 **
**7 11 14 18 22 25 29 32 **

**7 10 14 17 21 24 27 31 **
**6 10 13 16 19 23 26 29 **
**6 9 12 1518 21 24 27 **
**6 8 11 14 17 19 22 25 **
**5 8 10 13 15 18 20 23 **

**5 7 9 12 14 16 18 21 **
**4 6 8 10 12 14 16 19 **
**4 5 7 9 11 13 15 16 **
**3 5 6 8 10 11 13 14 **
**3 4 6 7 8 10 11 13 **

**2 4 5 6 7 8 10 11 **
**2 3 4 5 **

**I\**

**7 8 9**

**2 3 3 4**

### 5

**6 7 8**

**1 2 3 4 4 5 6 6**

**1 2 2 3 4 4 5 5**

**1 1 2 2 3 3 4 4 **
**1 1 2 2 2 3 3 4 **
**1 1 1 2 2 2 3 3 **
**0 I 1 1 1 2 2 2 2 **
**o 1 1 1 1 1 2 2 **

**o o 1 1 1 1 1 1 **
**o o o 1 1 **

### 1

**1 1**

**o o o o 1 1 1 1**

**o o o o o o o 1**

**o o o o o o o 1**

**o o o o o o o o **
**o o o o o o o o **
**o o o o o o o o **
**o o o o o o o o **
**o o o o o o o o **

**o o o o o o o o **
**o o o o o o o o **
**o o o o o o o o **

**1.2816 **

**0.990 **

**2.3263**

**1.3408 **

**0.991 **

**2.3656**

**1.4051 **

**0.992 **

**2.4089**

**1.4758 **

**0.993 **

**2.4573**

**1.5548 **

**0.994 **

**2.5121**

**1.6449 **

**0.995 **

**2.5758**

**1.7507 **

**0.996 **

**2.6521**

**1.8808 **

**0.997 **

**2.7478**

**1.9600 **

**0.998 **

**2.8782**

**6.3 Normal approximation to binomial distribution **

### When

*n *

### is fairly large (30 or more), andp is not too small (not less than about 0.1) or not

### too large (not greater than about 0.9), the

*normal *

### distribution with

µ,**= np **

**= np**

### and standard

### deviation a

**= **

### -Jnjiq can be used as an approximation for the binomial distribution.

### The histograms below and on the next page have been drawn for p = 0.8 and for values

### of

**n = **

**n =**

### 5, 10, 15 and 20. Observe that, as

*n *

### increases, the histograms become less skewed,

### leading then to the idea that the curve drawn through the midpoints of the top of each

### rectangle of the histogram has the characteristic symmetrical shape of the normal

### distribution curve.

### Pr

0.4

**0.3 **

**0.2 **

0.1

### I

**µ **

0

0

**2 **

3 4
Number of successes

Figure6-13:p

### =

*0.8,n*

### =

5,p.### =

4### Pr

**0.3 **

**0.2 **

0.1

### µ

0

**0 ** 2 3 4 5 6 7 8 9 10

Number of successes

### Pr

**0.3 **

**0.2 **

**0.1 **

**3 ** **6 7 8 9 10 11 12 13 14 15 **

**Number of successes **

**Figure 6-15: ****p ****= 0.8, n = 15, µ, = 12 **

### Pr

**0.3 **

**0.2 **

**0.1 **

**0 **

*I *
*I *

**µ **

**4 ** **8 ** **10 11 12 13 14 15 16 17 18 19 20 **
**Number of successes **

**Figure 6-16: p = 0.8, ****n ****= 20, µ, = 16 **

### The binomial variable is discrete and can assume only integral values 0,

**1, **

### 2, ... , n. How

### then can we represent its probability distribution by means of a histogram which can be

### drawn for a continuous variable only? This can be justified by the fact that the areas of

### the rectangles are proportional to the probabilities and, since the width of the base of each

### rectangle is one unit, the height of each rectangle represents the probability of the midpoint

### of the base. The sum of the areas of all the rectangles is 1 unit of area, corresponding to

### the fact that the sum of the probabilities is 1.

**Example 5 **

### A Gallup Poll establishes that 80 per cent of people interviewed are in favour of a certain

### proposal. If 20 people are interviewed, find, using the normal approximation to the

### binomial distribution, the probability that:

*µ*

**= **

**np **

**np**

**= **

### 20 X 0. 8

**= **

### 16

**a**

**a**

**= **

*-,/npq *

**= **

### --J16 X 0.2

**= **

### 1.789

### A normal distribution with mean 16 and standard deviation -13.2 can be used as

### an approximation in this case.

### a With reference to Figure 6-16 it will be seen that, to find the probability that

**X**

**X**

**= **

### 14,

**it **

### will be necessary to find the probability that 13.5

**<X* **

**<X***

### < 14.5. It

### should be remembered that the normal variable is continuous, whereas the

### binomial variable is discrete.

**Note: **

**Note:**

**If **

### Xis a binomial variable whose distribution is approximated by a normal

### variable

**X*, **

**X*,**

### then:

**Pr(X =a)""' Pr(a **

**Pr(X =a)""' Pr(a**

### -

**0.5<X*<a **

**0.5<X*<a**

### + 0.5)

### and:

**Pr(a <X **

**Pr(a <X**

### <

**b) **

**b)**

### ""'

**Pr(a **

**Pr(a**

**+ **

### 0.5

**<X* **

**<X***

### <

**b **

**b**

### - 0.5)

### for integer values

**a **

**a**

### and

*b *

### such that

**a< **

**a<**

*b.*

### WhenX*

**= **

### 13.5:

### WhenX*

**= **

### 14.5:

### z

**= **

### 13.5 - 16

""'### -1.40

### -13.2

### z

**= **

### 14.5 - 16""' -0.84

### -13.2

**Pr(X* **

**Pr(X***

### � 13.5)

**= **

**Pr(z � **

**Pr(z �**

### -1.40)

**= **

### 0.0808

**Pr(X* **

**Pr(X***

### < 14.5)

**= **

**Pr(z **

**Pr(z**

### < -0.84)

**= **

### 0.2005

### Pr(13.5 <

**X* **

**X***

### < 14.5)

**= **

### 0.2005 - 0.0808

**= **

### 0.1197

### Check whether this is a good approximation by evaluating

_{(i�) }

### (0.2)

**6**

_{(0.8) }

_{14• }### b It will be necessary to find Pr(14.5

**<X* **

**<X***

### < 17.5)

### When

**X* **

**X***

**= **

### 14.5:

### WhenX*

**= **

### 17.5:

### z

**= **

### 14.5 - 16

""'### -0.84

### -13.2

### z

**= **

*11.5 *

### - 16

""'### 0.84

### -13.2

**Pr(X* **

**Pr(X***

### � 14.5)

**= **

**Pr(z � **

**Pr(z �**

### -0.84)

**= **

### 0.2005

**Pr(X*< **

**Pr(X*<**

### 17.5)

**= **

**Pr(z< **

**Pr(z<**

### 0.84)

**= **

### 0.7995

### Pr(14.5<X*< 17.5)

**= **

### 0.7995 - 0.2005

**= **

### 0.5990

### Compare this result with the arithmetical drudgery involved in calculating:

**Exercises 6b **

### (In each of the following questions, use the normal approximation to the binomial

### distribution where applicable.)

### 1 A dental inspector finds that about 20 per cent of children of a certain area have tooth

### decay. If a group of 400 children is randomly selected:

**a **

### how many would be expected to have tooth decay?

**b **

### what is the probability of exactly this number?

**c **

### what is the probability that the number lies within one standard deviation of the

### expected number?

**2 **

### A fair coin is tossed 100 times.

### a How many heads do we expect to turn up?

**b **

### What is the probability of this number?

### c What is the probability that the number of heads is greater than 45 but less than 55?

**3 **

### A fair coin is tossed 500 times. Find the probability that the number of heads uppermost

### will not differ from 250 by more than 10.

**4 **

### A fair die is thrown 180 times. What is the probability that:

**a **

### a six will turn up exactly 40 times?

### b an odd number will turn up at least 100 times?

**5 **

### A targetshooter finds that a bull's-eye is scored on 20 per cent of occasions. What is

### the probability that at least 24 bull's-eyes will be scored out of 100 attempts?

**6 **

**a **

### Assuming that the length of life of a certain type of television tube is normally

### distributed with a mean of 1000 hours and a standard deviation of 250 hours, what

### proportion of tubes would be expected to have a life not exceeding 780 hours?

**b **

**If 100 such tubes are randomly selected, how many would be expected to have a life **

### not exceeding 780 hours and what is the probability that the number exceeds 21?

**7 **

### A manufacturer of metal pistons finds, that on average, 10 per cent of the pistons are

### rejected because they are either oversize or undersize. What is the probability that a

### batch of 900 pistons will contain:

**a **

### no more than 100 rejects?

**b **

### at least 80 rejects?

**8 **

### Hospital records show that of patients suffering from a certain complaint, 75 per cent

### recover. What is the probability that, of 48 randomly selected patients, at least 40

### recover?

**9 **

### In packets of flower seeds, 40 per cent are known to produce pink flowers. If 250 seeds

### are planted and they all flower:

**a **

### how many pink flowers would we expect?

**b **

### what is the probability of this number?

**c **

### within what limits·would the number of pink flowers very probably lie?

**6.4 Probability limits for a single value of the **

**normal /variable **

### It has been stated that one of the characteristic properties of a normal distribution is that:

**(i)**

### about� of the population lies in the intervalµ ±

*a *

### (Figure 6-17).

**68.27% **

### µ-cr--µ µ+cr

**Figure 6-17 **

**95.45% ** **99.74% **

### µ -20' µ µ + 20' µ-30' µ µ + 30'

**Figure 6-18 ** **Figure 6-19 **

### From this we infer that a single value,

**x, **

**x,**

### of the variable will

**almost certainly **

**almost certainly**

### lie within 3

### standard deviations of the mean:

### i.e.

### I almost certainly,

µ,### -

*3a � x � *

### µ, +

*3a.*

**Very probably **

**Very probably**

### (probability of about 0.95) it will lie within 2 standard deviations of the

### mean:

### i.e.

### very probably,

µ,### -

*2a � x � *

### µ, +

*2a.*

### These limits for the value of a variable are called the 3 sigma and 2 sigma

**probability limits**

**probability limits**

### respectively. If a value of the variable lies beyond either of these limits, it is said to differ

**significantly **

**significantly**

### from the mean at that particular level of significance.

**Example& **

### A manufacturer of electric light globes finds that the globes have an average life of 2000

### burning hours with a standard deviation of 200 hours. Assuming that the distribution of

### lifetimes is normal, within what interval will the lifetimes almost certainly lie?

### Their lifetimes will almost certainly lie in the interval

µ,### ±

*3a.*

### i.e.

### in the interval 2000 ± 600 hours

### i.e.

### between 1400 and 2600 hours

### If a globe, randomly selected, had a lifetime of only 1200 hours, what conclusion could we

### draw?

### We could conclude thit this is most unlikely to have occurred by chance so perhaps there

### is some factor which needs to be taken into consideration, such as a fault in the

### manufacturing process.

### Further sampling would be necessary to discover whether the lifetime of globes was

### consistently lower than expected.

**Example 7 **

### In the long run, 64 per cent of patients treated for a particular disease with drug

*X *

### are

### cured. If 100 patients, not specially selected, are treated with this drug and 75 are cured,

### determine whether this number is significantly higher than the expected numbe� of cures.

### Expected number of cures:

### µ

**= **

*np *

### =

### 100 X 0.64

### = 64

*a *

**= **

*-Jnpq *

### =

*-JlO0 *

### X 0.64 X 0.36

### = 4.8

### µ

### +

*2a *

**= **

### 64

### +

### 2 X 4.8

**= **

### 73.6

### µ

### +

*3a *

### = 64

### +

### 3 X 4.8

### =

### 78.4

### So 75 cures are significantly more than the expected number at the 2 sigma limit

### but not at the 3 sigma limit.

### The mean, µ, is a characteristic parameter of a Poisson distribution and the variance and

### standard deviation areµ and -Jµ, respectively, the derivation of which is beyond the scope

### of this book. Whenµ, is large, the normal distribution gives a satisfactory approximation

### to the Poisson distribution with practically all values of the Poisson variable lying in the

### rangeµ ± 3-Jµ, since a

**= **

### -Jµ, and very probably (probability of about 0.95) lying in the

### range µ ± 2-Jµ,.

**Example& **

### The number of demands for a certain item of equipment varies randomly from week to

### week, following a Poisson distribution with mean 20. What is the smallest number of items

### a firm must have in stock each week to be almost certain of not having to refuse a demand

### for this item?

### µ

### =

### 20

*a*

### = ,,/µ

**= .J26 = **

### 4 .4 72.

### It is almost certain that the demand will not be more thanµ, + 3-Jµ,, i.e. not more

### than 20

### +

### 3 x 4.472

### = 33.416.

### The firm should have at least 34 items in stock.

### Very probably the demand will lie in the interv.al µ, ± 2-Jµ,, i.e. in the interval

### 20 ± 2 x 4.472. The demand will very probably be between 11 and 29.

**Exercises 6c **

### 1 Assuming that the length of life of a certain type of television tube is normally

### distributed with a mean of 1000 hours and standard deviation of 250 hours, after how

### many hours is a tube almost certain to fail?

### NORMAL DISTRIBUTION 185

### 3 A dental inspector finds that about 20 per cent of children of a certain age have tooth

### decay. In a certain area the inspector finds that 25 out of 200 children examined have

### tooth decay. Use 3 sigma limits to determine whether this number is significantly less

### than the expected number.

### 4 A Gallup Poll establishes that 60 per cent of people are in favour of a certain proposal.

### If a sample of 120 were interviewed, use 2 sigma limits to estimate the number in favour

### of the proposal.

### 5 Electricity power failures occur according to a Poisson law with an average of three

### failures every twenty weeks. If, over a period of 40 weeks, there were actually 9 failures,

### use 3 sigma limits to determine whether this is significantly more than the expected

### number.

**6 **

### Variables

**X **

**X**

### and

**Y **

**Y**

### are known to be connected by the formula:

**Y **

**Y**

### = 10 +

**bX.**

**bX.**

**X **

**X**

### can be measured accurately but the measurements of

**Y **

**Y**

### are subject to a random error

### which is normally distributed with mean of 0 and standard deviation of 0.20. An

### observation

**Y **

**Y**

### = 15 is obtained when

**X **

**X**

### = 2. Determine limits within which

**b **

**b**

### almost

### certainly lies.

### 7 The cost,

*$C *

### per article, of manufacturing an article is related to the weight,

*w *

### g, by

### the equation:

*C *

### =

*2w *

### + 25

### The weight of the articles is normally distributed with mean_ 5 g and standard deviation

### 0.1 g. Give limits within which the cost of the article will almost certainly lie.

**8 **

### On the average, one student in every ten wears glasses.

### a From a group of 90 students, how many would be expected to wear glasses? Give

### limits between which this number very probably lies.

### b How large would a group of such students need to be for us to be almost certain

### that the number of students in the group wearing glasses is at least 63?

**9 **

### A fair coin is tossed 100 times.

### a What is the mean and standard deviation of the number of heads appearing

### uppermost?

### b Give limits between which the number of heads:

**(i)**

### very probably will lie

**(ii)**

### almost certainly will lie.

**10 A **

### manufacturer of metal pistons markets the product in batches of 10 and finds that

### 15 per cent of batches contain at least one defective piston. In 1000 batches, estimate

### the mean and standard deviation of the number of defective pistons and give limits

### between which this number will almost certainly lie, assuming the Poisson law.

**11 **

### The number of demands for a certain item of equipment varies randomly from week

### to week, following a Poisson distribution with mean 4. If

**X **

**X**

### denotes the number of

### demands per week, find

**Pr(X � **

**Pr(X �**

### µ, - a).

**12 **

### A retailer keeps a record of sales and finds that, on 82 out of 1000 days, there was no

### demand for a particular item of clothing. On one particular day there was a demand

### for 13 such items. Assuming the Poisson law, determine whether this is significantly

### more than the expected number.

**13 **

### Electricity power failures occur according to a Poisson law, with an average of three

### failures every twenty weeks. If, over a period of 40 weeks, there were actually nine

### failures, use three sigma limits to determine whether this is significantly more than the

### expected number.

*·/*

### 14 Cans of peas are tested for infection by certain organisms by storing them for a period

### of time before they leave the factory. Cans which contain one or more organisms burst

### open on account of fermentation.

### a

**If **

### the number of organisms in a can is a Poisson variate, and on the average 76 in

### every 10 000 cans burst open, find the mean number of organisms per can.

### b In batches of 5000 cans, find approximately the mean and standard deviation of the

### number of cans which burst open, and give limits between which this number will

### almost certainly lie.

### 15 In samples of milk taken from a bulk transportation vehicle, 40 per cent proved to have

### no bacterial spores.

### a Assuming the Poisson law, estimate the mean number of bacterial spores per sample

### and determine the proportion of samples-which would contain t�o bacterial spores.

### b Out of 1000 samples, how many would be expected to have only-one spore each?

### Give limits between which this number very probably lies,

**6.5 Probability limits for the sample mean of n**

**6.5 Probability limits for the sample mean of n**

**values of the variable **

**If **

### a random sample of

*n *

### observations is drawn from a normally distributed population with

### mean

µ,### and standard deviation

*<J,*

### it can be shown that the mean,

*x, *

### of the sample:

/

**(i)**

*more likely than not*

### (probability about�) lies in the interval

µ,### ±

*Jn *

### (Figure 6-20).

**(ii)**

*very probably*

### (probability of about 0.95) lies in the interval

µ,### ±

### 2

*J,, *

### (Figure 6-21).

*n *

### i.e.

### very probably,µ, - 2

*J,, :,;; *

### x:,;; µ, + 2

*J,, *

**(iii)**

*almost certainly*

### lies in the interval

µ,### ±

### 3

*Jn *

### (Figure 6-22).

### i.e.

_{almost certainly, µ, - 3 }

*Jn :,;; *

_{x :,;; µ, ,+ 3 }

*Jn *

68.27%

### µ-.Q. µ µ+.Q.

*.Jn *

*.Jn *

Figure 6'...20

95.45% 99.74%

### µ-gg µ µ+gg µ-;m µ µ+�

*.Jn *

*.Jn *

*.Jn *

*.Jn *

### The quantity

*J,, *

### is known as the

**standard error **

**standard error**

### of the mean. So, in Example 6, the mean

### life of 100 randomly selected globes would

**almost certainly **

**almost certainly**

### lie in the interval

### 2000

**± **

### 3

*:;..:oo, *

### i.e. in the interval 2000 ± 60 h, and

**very probably **

**very probably**

### would lie in the interval

### 100

### 2000

**± **

### 2

_{�}

_{100 }

### O, i.e. in the interval 2000 ± 40 h.

### The standard error, J,,, gets smaller and smaller as n gets larger and larger, and

### J,,---+O asn---+oo.

### n

### If

**n **

**n**

### = 400, the mean life of 400 randomly selected globes would almost certainly lie in the

### interval 2000 ± 3 x �. i.e. in the interval 2000 ± 30.

### -v400

### As

**n **

**n**

### increases, x should give us a more reliable estimate of

µ,. This is what we would expect.### It seems feasible to suggest, then, that in order to get a true estimate of the population

### mean, take as large a sample as possible. However, this is not practical in many situations ..

### Certainly, in the case of the globes it would be wasteful and expensive. Why?

### We use general phrases such as 'more likely than not', 'very probably', 'almost certainly'

### when referring to probability limits and levels of significance such as the 2 sigma level and

### the 3 sigma level. The probability that x lies in the intervalµ, ± 1.96 � is 0.95.

_{. }

_{. }**-vn **

**-vn**

**Example 9 **

### The mean weekly wage in a c,ertain industry is $500 with a standard deviation of $30. A

### random sample of 25 employees in this industry has a mean wage of $475. ls this

### significantly less, at the 3 sigma level, than the mean wage of the population?

### At the 3 sigma level,

### µ,

**± **

### 3 �

**= **

### 500

**± **

### 3 X �

**. **

**.**

**-vn **

**-vn**

### -v25

### = 500 ± 18

### Since 475 is not in this interval, it is significantly less at the 3 sigma level.

**Exercises 6d **

### 1 The mean weight of boys of a certain age is 50 kilograms with standard deviation of 5

### kilograms. Within what limits would the mean weight of a random sample of 64 boys

### of this age very probably be?

### 2 At a certain school, the mean IQ (Intelligence Quotient) of the students is 100, with a

### standard deviation of 15. The mean IQ of a sample of 25 students was 112. Is this

### significantly higher than would be expected?

### 3 A sample, 55, 63,--69;-'B, is drawn from a population whose standard deviation is 4 and

### whose mean is thought to be 60. Do you think the population mean has been wrongly

### given?

### -�

### 4 A machine makes electrical resistors which have a mean resistance of 50 ohms with a

### standard deviation of 2 ohms.

### a Within what limits would we expe�t the mean of 25 randomly selected resistors to

### lie with a probability of 0.95? W\'',

*0*

_{' }### 5 Butter is marketed to retailers in cartons containing 16 packages drawn from a

### population normally distributed with a mean weight of 0.5 kg and a standard deviation

### of 0.02 kg.The butter in a particular carton weighed 8.15 kg. Is this significantly more

### than the expected weight at the 2 sigma level of significance?

### 6 The mean height of a sample of 25 students is 150 cm. Can we infer that this sample is

### drawn from a population of students of mean height 160 cm and standard deviation

### 10 cm?

_{'1}

### -:: 2

### �

### 7 The length of a certain species of fish has a normal distribution with mean 30 cm and

### standard deviation 2.5 cm. An angler caught nine such fish whose average length was

### 27 cm. Is this significantly less than the expected value at the

*3a *

### level?

**6.6 Confidence limits **

**a Population mean **

### Confidence limits are limits for the value of a parameter estimated from a particular value

### of a statistic. Discussion will be confined in this section to estimating a population mean

### from a single value of the variable or from the mean of a set of

*n *

### observations.

**If **

*xis *

### a particular value of a variable, it was stated in Section 6.4 that almost certainly:

*µ, *

### -

*3a *

### ,,;;

*x *

### ,,;;

*µ, *

### +

*3a *

### . . . .. . .

**(1)**

### Transposing

**(1) **

### gives:

*µ, *

### ,,;;

*x *

### +

*3a *

### and

*x *

### -

*3a *

### ,,;;

*µ,*

### i.e.

*x *

### -

*3a *

### ,,;;

*µ, *

### ,,;;

*x *

### +

*3a *

### ...

**(2)**

### The population mean,

*µ,, *

### therefore almost certainly lies in the interval x

**± **

*3a. *

### These limits

### for the value of

µ, are called the 3 sigma confidence limits. Very probably,*µ, *

### lies between

*x *

**± **

*2a, *

### these limits being called the 2 sigma confidence limits.

**If **

### we have a

*random sample of n observations *

### with mean

*x, *

### we would expect

*x *

### to give a

### better estimate of

µ,### than

*a single value *

### of the variable.

### It can be shown that

*µ, *

### almost certainly lies in the interval x

**± **

### 3;. The standard error, �.

### will decrease as

*n *

### increases.

'V*n *

'V *n *

### We have assumed that the standard deviation,

*a, *

### of the population is known.

**If **

### it is not

### known, the sample standard deviation,

*s, *

### may be used as an estimate of it if the sample

### is large.

### The approximate 95% confidence interval for

µ, would be given by:**Example 10 **

*-*

*2a*

*-*

*2a*

*x- c,,;;µ,,,;;x+*

_{vn ·.}

_{vn ·.}*---,*

_{vn }

_{vn }### The IQ (Intelligence Quotient) of a sample of 100 VCE students had a mean of 108 with

### a standard deviation of 15. Find a 95% confidence interval for the mean IQ of the

### population of VCE students.

*n *

**= **

### 100

### x

**= **

### 108

*s *

**= **

### 15

**= **

### estimate of

*a*

*ax *

**= **

### __!!__

**= **

### __11_

**= **

### 1 5

### The approximate 95% confidence interval would be given by:

### -

### 2a

### -

### 2a

### x- -Jn �µ�x+ -Jn

### i.e.

### 108 - 2 X 1.5 � µ � 108 + 2 X 1.5

### 105 � µ � 111

### We can be about 95 % sure that the mean IQ of the population lies in the interval

### 105 to 111. The confidence limits are 105 and 111.

**Example 11 **

### A random sample of 25 employees has a mean weekly wage of $320. Could this sample have

### been taken from a population of employees whose weekly wage is normally distributed with

### mean of $290 and standard deviation of $40?

**n **

**n**

**= **

### 25

### x =

### 320

### a= 40

### ax=_.!!_= 40

**= **

### 8

### -Jn 5

### We can be almost certain that the mean wage of the population would be in the

### interval .x ±

*;fn *

### i.e. in the closed interval 320 ± 3 x 8, i.e. (296, 344]

### Since 290 does not lie in this interval, we can reject the hypothesis that the sample

### is taken from a population whose mean weekly wage is $290.

**b Proportions **

**Example 12 **

### A random sample of 400 manufactured articles contains 80 defectives. Give 2 sigma

### confidence limits for the number of defectives in samples of 400 articles and give the

### proportion of defectives in the whole output of all samples of 400 articles.

### Using this sample to estimate the probability of defectives:

**n = **

**n =**

### 400

_{p = 400 = 0·2 }

### -

### 80

### where j3 denotes the

*sample *

### proportion.

### q

**= **

### 0.8

### Standard deviation

**=**

### -J npq

**= v**

r-### 4

--,-### 00

-,---### x-0

----,### . 2

,---### x-o

-=--### .----=8

**= **

### 8.

### The 2 sigma confidence limits for the number of defectives in samples of 400 articles

### are 80

**± **

### 2 x 8. At this confidence level, the number of defectives lies between 64

### and 96. Therefore the proportion of defectives lies between 4

### �6

### and

*1*

_{0}

### 6

_{0}

_{,}

*Alternatively: *

### Usingp as the sample proportion:

### -

**X**### 80 O

### P = n = 400 = ·2

### q = l - p = 0.8

### The 95% confidence intervals for pare given by:

### P - 2 �P (1 n-p) :,;;,_ P :,;;,_ P + 2 �P (1 n-p)

### i.e.

_{0.2 _ 2. /0.24�0}

_{'\J }

### 0.8 :5:: :5:: 0 2 + 2. /0.2 X 0.8

### ..._,_p ..._,_ •

_{'\J 400 }

### i.e.

_{0.16 :,;;,_ }

*p *

### :,;;,_ 0.24 as before

**Formulae for the mean and standard deviation of a binomial **

**distribution **

*Random variable *

*Mean *

### Number of occurrences

*(np) *

_{µ}

_{µ}

_{njj }### =

_{np }

_{np }

### Proportion of occurrences (p)

*µ. *

_{p }_{=P }

_{=P }

**Example 13 **

*Standard deviation *

*<l*

*n;;*

### =

*.Jnpq *

*<l*

*;;*

*=*

*vnpq=*

_{n }

_{n }

*�*

_{n }

_{n }