Chapter III. Probability. 2015.doc

Chapter III

Probability Concepts and

Applications

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Objetive

Chapter

1.1 Introduction

A probability is a numerical statement about likelihood that an event will occur. In this chapter we develop probability models that can be used to study business for which future outcomes are unknown.

A principal objective of statistics is inferential statistics, which is to infer or make logical decisions about situations or population simply by taking and measuring the data from a sample. This sample is taking from a population, which is the entire group in which we are interested. We used the information from this sample to infer conclusions about the population.

For example, we are interested to know how people will vote in a certain election. We sample the opinion of 1500 of the electorate and we used this result to estimate the opinion of the population of 11 million. Since we are extending our sample results beyond the data that we have measured, this means that there is no guarantee but only a probability of being correct or of making the right decision. The corollary to this is that there is a probability or risk of being incorrect.

Other areas of business:

 Quality control: sampling schemes based on probability, e.g. acceptance sampling.  Sample design for sample surveys.

 In decision making, e.g. the use of decision trees.

 Stock control, e.g. probable demand for stock before delivery of next order.

Probability can be calculated from various sources:  Simple probability calculated from symmetry;

 Simple probability calculated from frequencies;

 Conditional probability calculated from contingency tables;  Conditional probability calculated from tree diagrams; etc.

1.2 Concept of probability

The concept of probability is the chance that something happens or will not happen. In statistics it is denoted by the capital letter P and is measured on an inclusive numerical scale of 0 ≤ P(event) ≤ 1. If we are using percentages, then the scale is from 0% to 100%. If the probability is 0%, then there is absolutely no chance that an outcome will occur. For example I live in Rwanda, but I born in Peru, the probability of I becoming president is 0%. At the top end the probability scale is 100% which means that it is certain the outcome will occur. The higher the degree of probability, the more likely the event is to happen, or, in a longer series of samples, the greater the number of times such event is expected to happen.

Random Experiment

A random experiment is a process leading to two or more possible outcomes, with uncertainly as to which outcome will occur.

Examples:

1. A customer enters a store and either purchases a shirt or does not. 2. The daily change in an index of stock market prices is observed.

3. A bad cereal is selected from a packing line and weighed to determine if the weight is above or below the stated package weight.

4. Some number of persons will be admitted to a hospital emergency room during any hour.

Sample Space: W

A sample space W for an experiment is the set for all possible outcomes of the experiment. The elements of W are called simple points

Example:

What is the sample space for the roll of a single six-sided die?

Solution: The basic outcomes are the six possible face numbers, and the sample space is: W = {1,2,3,4,5,6} Þn(W) = 6

The sample space contains six basic outcomes. No two outcomes can occur together, and one of the six must occur.

Note: In many cases we are interested in some subset of the basic outcomes and no the individual outcomes. For example, for the roll of a die we might be interested in whether the outcome is even – that is, 2, 4, or 6.

Events

An Event, E, is any subset of basic outcomes from the sample space. An event occurs if the random experiments results in one of its constituent basic outcomes. The null event represents the absence of a basic outcome and is denoted by Ø.

Example: A die is rolled, W = {1,2,3,4,5,6}, Let A be the event “Number resulting is even” and B the event “Number resulting is at least 4”. Then:

A= {2,4,6}; and B= {4,5,6}

Find the complement of each event, the intersection and the union of A and B, and the intersection of and B.

The complements of these events are, respectively, = {1,3,5}; and = {1,2,3}

The intersection of A and B is the event “Number resulting is both even and at least 4” and so

The union of A and B is the event “Number resulting is either even or at least 4, or both” and so Intersection of and B

Probability is measure of uncertainty

Probability measures the extent to which an event is likely to occur and can be calculated from the ratio of favorable outcomes to the whole number of all possible outcomes.

There are two main ways of assessing the probability of a single outcome occurring:  from the symmetry of a situation to give an intuitive expectation,

 from past experience using relative frequencies to calculate present or future probability. Value of Probability

Probability always has a value between 0, impossibility, and 1, certainty. Probability of 0.0 corresponds to ‘impossible’

0.1 corresponds to ‘extremely unlikely’ 0.5 corresponds to ‘evens chance’ 0.8 corresponds to ‘very likely’ 1.0 corresponds to ‘certainty’

Probability from symmetry

We can define the probability of an event A taking place as:

Invoking symmetry there is no need to actually do any experiments. This is the basis for the theory of probability which was developed for use in the gaming situation. – 'a priori' probability.

Examples

1. Tossing 1 coin:

Possibilities: (Head, Tail). P(a head) = 1/2 = 0.5 P(a tail) = 1/2 = 0.5 2. Tossing 2 coins:

Possibilities: Ω= (Head Head, Head Tail, Tail Head, Tail Tail.) a. Two Heats occurs

If E= (H H), the E is a simple event, so

b. At least one Heat occurs: If F= (HH, HT, TH)

Which has three outcomes. Thus,

Note that the sum of all the possible probabilities is always one. This makes sense as one of them must occur and P(certainty) = 1. 3. Rolling a pair of dice

2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6

Find the probability of getting a sum of 8 points by throwing a die Answer: 5/36

Questions:

P(double six)= P(any double)= P(sum less than 4)= P(more than 8)= P(at least one six)=

Probability from Frequency

An alternative way to look at Probability would be to carry out an experiment to determine the proportion of favorable outcomes in the long run.

→ Number of occurrences of the event

→ Total number of trials or outcomes

This is the frequency definition of Probability.

The relative frequency of an event is simply the proportion of the possible times it has occurred.

The relative frequency probability is the limit of the proportion of times that event A occurs in a large number of trials, n:

Where nA is the number of A outcomes and n is the total number of trials or outcomes.

The probability is the limit as n becomes large (or approaches infinity). Example:

Niyomwungeri Lydiane is considering an opportunity to establish a new car dealership in Kigali City, which has a population of 150,000 people. Experience from many other dealerships indicates than in similar areas dealerships will be successful if at least 40% of the households have annual incomes over 750000 Rwf. She has asked Shingiro Eric, a marketing consultant, to estimate the proportion of family incomes above 750000, or the probability of such incomes.

Solution: After considering the problem, Eric decides that the probability should be based on the relative frequency. He first examines the most recent census data and finds that there were 54,345 households in Kigaly City and that 11,496 had incomes above 750000. Eric computed the probability for event A, “Family income greater than 750000 Rwf” as

Since Eric knows that there are various errors in census data, he also consulted similar data published by Sales Management Magazine. From this source he found 55,100 households, with 11500 having incomes above 750000 Rwf. Eric computed the probability of event A from this source as

1. P(A) ³ 0 2. 0 ≤ P(A)≥ 1 3. P(W) = 1

4. If A and B are mutually exclusive: P(A U B) = P(A)+P(B), in general: P(A1 U A2 U.... U An) = P(A1)+P(A2)+....+P(An)

5. Adding Probability of Events

a) If A, and B are not mutually exclusive P(A U B) = P(A)+P(B) - P(A ∩ B)

b) If A, B and C are not mutually exclusive P(A U B U C) =

P(A U B U C) = P(A)+P(B)+P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B∩ C)

Contingency tables

Contingency tables display all possible combined outcomes and the frequency with which each of them has happened in the past. These frequencies are used to calculate future probabilities.

Example. A new supermarket is to be built in City Town. In order to estimate the requirements of the local community, a survey was carried out with a similar community in a neighboring area. Parts of the results of this are summarized in the table below:

Expenditure on sweets

Mode of travel None Under 20 Rwf At least 20 Rwf Total

On foot 40 20 10 70

By bus 30 35 15 80

By car 25 33 42 100

Total 95 88 67 250

Suppose we put all invoices into a drum and thoroughly mix them up. If we close our eyes and take out one invoice, we have selected one customer at random. Calculate:

a. P(customer spends over Rwf 20) = 67/250 = 0.268 b. P(customer will travel by car) =100/250=0.40

From the ‘cells’ we can calculate, for example:

c. P(customer spends over Rwf 20 and travels by car) = 42/250 = 0.168

d. P(customer arrives on foot or spends no money on sweets) = (40 + 20 + 10 + 30 + 25)/250 =0.50 or (70+95-40)/250= 0.50

Sometimes we need to select more than one row or column:

f. P(a customer will spend less than Rwf 20) = (95 + 88)/250 = 0.732 g. P(a customer will not travel by car) =

In the examples above all the customers have been under consideration without any condition being applied which might exclude any of them from the overall ratio. Often not all are included as some condition applies.

Conditional Probability

If the probability of the outcome of a second event depends upon the outcome of a previous event then the second is conditional on the result of the first. This does not imply a time sequence but simply that we are asked to find the probability of an event given additional information - an extra condition, i.e. if we know that the customer travelled by car all the other customers who did not are excluded from the calculations.

For example: P(a customer spends at least Rwf 20, if it is known that he/she travelled by car)

We eliminate from the choice all those who did not arrive in a car, i.e. we are only interested in the third row of the table. A short hand method of writing 'if it is know that. .' or 'given that . . .' is‘.

a. P(a customer spends at least Rwf 20, | he/she travelled by car) = 42/100 = 0.420 b. P(a customer came by car | he/she spent at least Rwf 20 on sweets) = 42/67 = 0.627 Note that

P(spending ³ Rwf 20, | he/she travelled by car) ¹ P(coming by car | he spent ³ Rwf 20 on sweets) c. P(a customer spends at least Rwf 20, | he/she travelled by bus) =

d. P(a customer came on foot, | he/she spent at least Rwf 20 on sweets) =

E

Employment

Sex Unemployed Employee Total

Men 40 460 500

Women 260 140 400

Total 300 600 900

Adult is chosen randomly, what is the probability that: a. Be unemployed

b. Be unemployed as a woman? c. Be unemployed as a men? SOLUTION:

b.

Interpretation: When choosing an adult at random, the probability that the unemployed as a woman is 0.65.

Is a variable whose value is a number determined by the outcome of a random experiment is denoted by capital letters: X, Y, Z. and their individual values by the corresponding lowercase letter, if X (is a random variable) is, then their values are: X1, X2,... Xn

If every value of the random variable its probability we associate obtain the probability distribution of that variable.

Example: Let an experiment of tossing 2 coins and let us note the results of the head.

We define the random variable “Y” as the number of heads obtained in the 2 releases. Thus, taking the variable values may be 0, 1, 2 depending on the results of the experiment.

Sample space

Random Variable

TT 0

HT 1

TH 1

HH ₂

# Ω=4

Then the probability distribution of this variable will be

Random

Variable _Probability

0 1/4

1 2/4

2 _1/4

Total 1

EXPECTED VALUES

A commonly used method in decision making problems is the consideration of expected values. The expected value for each decision is calculated and the option with the maximum or minimum value (dependent upon the situation) is selected.

The expected value of each decision is defined by:

E(x) = åpx

Example: Two independent operations A and B are started simultaneously. The times for the operations are uncertain, with the probabilities given below:

Operation A Operation B

Duration (days) (x)

Probability (p)

Duration (days) (x)

Probability (p)

1 0 1 0.1

2 0.5 2 0.2

3 0.3 3 0.5

4 0.2 4 0.2

Operation A

E(x) = åpx = (1 x 0.0) + (2 x 0.5) + (3 x 0.3) + (4 x 0.2) = 2.7 days Operation B

E(x) = åpx = (1 x 0.1) + (2 x 0.2) + (3 x 0.5) + (4 x 0.2) = 2.8 days Hence Operation A has the shorter completion time.

PROBABILITY DISTRIBUTIONS

Probability distributions are a mathematical model which can be used to describe a set of data.

You should by now be able to recognize whether the data is discrete or continuous, and a brief look at a histogram should give you some idea about its shape. The choice of an appropriate distribution depends upon the 'shape' of the data and whether it is discrete or continuous.

Probability distribution for Continuous Random Variable: Normal Distribution

T Student Distribution F Snedecor Distribution Exponential Distribution Chi-Square Distribution

Probability distribution for Discrete Random Variable:

Binomial Distribution: yes/no experiments (two possible outcomes) Poisson Distribution

Hypergeometric Distribution

Normal Probability Distribution

The normal probability distribution is one of many which can be used to describe a set of data. The choice of an appropriate distribution depends upon the 'shape' of the data and whether it is discrete or continuous.

Where m and s2 _{are any numbers such that and where and physical}

constants,

The Normal distribution is found to be a suitable model for many naturally occurring variables which tend to be symmetrically distributed about a central modal value - the mean.

 Centered around the mean (

m

)

s

)

If the variable described is continuous, its probability function is described as a probability density function which always has a smooth curve under which the area enclosed is unity.

The Characteristics of any Normal Distribution

The normal distribution approximately fits the actual observed frequency distributions of many naturally occurring phenomena e.g. human characteristics such as height, weight, IQ etc. and also the output from many processes, e.g. weights, volumes, etc.

There is no single normal curve, but a family of curves, each one defined by its mean, µ, and standard deviation, s; µ and s are called the parameters of the distribution.

As we can see the curves may have different centers and/or different spreads but they all have certain characteristics in common:

 The curve is bell-shaped,

 It is symmetrical about the mean (µ),  The mean, mode and median coincide. The Area beneath the Normal Distribution Curve No matter what the values of µ and s are for a normal probability distribution, the total area under the curve is equal to one. We can therefore consider partial areas under the curve as representing probabilities.

The partial area between a stated numbers of standard deviation below and above the mean is always the same, as illustrated next. The exact figures for the whole

distribution are to be found in standard normal tables.

Note that the curve neither finishes nor meets the horizontal axis at m ± 3s, it only approaches it and actually goes on indefinitely.

The Standard Normal Distribution

• It is tedious to ‘integrate’ a new normal distribution for every population, so use a ‘standard normal distribution’ with standard tabulated areas.

• Convert your measurement x to a standard score (z-score): z = (x - m) / s

• Use the standard normal distribution: m = 0 and s = 1 (areas tabulated in any statistics text book or Excel)

No matter what units are used to measure the original data, the first step in any calculation is to transform the data so that it becomes a standardized normal variate following the standard distribution which has a mean of zero and a standard deviation of one. The effect of this transformation is to state how many standard deviations a particular given value is away from the mean. This standardized normal variate is without units as they cancel out in the calculation.

“The z-score indicates the number of standard deviation that value x is away from the mean µ”

Standard Normal Distribution

The formula for calculating the exact number of standard deviations (z) away from the mean (m) is: ;

The process of calculating z is known as standardizing, producing the standardized value which is usually denoted by the letter z, though some tables may differ.

Knowing the value of z enables us to find, from the Normal tables, the area under the curve between the given value (x) and the mean (m) therefore providing the probability of a value being found between these two values.

Finding Probabilities under a N

Normal Curve

The steps in the procedure are:  draw a sketch of the situation,

 standardize the value of interest, x, to give z,

 convert this to: a probability, using the area as found since total area = 1; or a percentage, multiplying the area by 100 since total area = 100%; or a frequency, multiplying the area by the total frequency; as required by the question.

How Do You Use The Normal Distribution? • Use the area UNDER the normal distribution

For example, the area under the curve between x=a and x=b is the probability that your next measurement of x will fall between a and b

Example: A normal distribution with a mean of 75 and a standard deviation of 10. The shaded area contains 95% of the area and extends from 55.4 to 94.6. For all normal distributions, 95% of the area is within 1.96 standard deviations of the mean.

How Do You Get m and s?

• To draw a normal distribution you must know m and s

• If you made an infinite number of measurements, their mean would be m and their standard deviation would be s

In practice, you have a finite number of measurements with mean x and standard deviation s • For now, m and s will be given; later we’ll use x and s to estimate m and s

Example

In order to estimate likely expenditure by customers at a new supermarket, a sample of till slips from a similar supermarket describing the weekly amounts spent by 500 randomly selected customers was analyzed. These data were found to be approximately normally distributed with a mean of Rwf 50 and a standard deviation of Rwf 15. Using this knowledge we can find the following information for shoppers at the new supermarket.

 The probability that any shopper selected at random: a) spends more than Rwf 80 per week,

b) spends less than Rwf 50 per week.

c) spend between Rwf 30 and Rwf 80 per week, d) spend between Rwf 55 and Rwf 70 per week.  The expected number of shoppers who will: e) spend less than Rwf 70 per week,

f) spend between Rwf 37.50 and Rwf 57.50 per week.  The value below which:

g) 70% of the customers are expected to spend, Solution

We assume that expenditure by customers is normally distributed. In this way, we use the Normal Distribution as a model for measurement.

First, we transform expenditure by customers into a Z-score, using the z-score transformation equation:

a) The probability that a shopper selected at random spends more than Rwf 80 per week

 Find P(Z > a). The probability that a standard normal random variable (z) is greater than a given value (a) is easy to find. The table shows the P(Z < a). The P(Z > a) = 1 - P(Z < a).

Suppose, for example, that we want to know the probability that a z-score will be greater than 3.00: From the Normal probabilistic table, we find that P(Z < 3.00) = 0.9987. Therefore, P(Z > 3.00) = 1 - P(Z < 3.00) = 1 - 0.9987 = 0.0013.

 In our example:

We need P(x > Rwf 80), the probability that a customer spends over Rwf 80

First standardize:

From tables: (z in the margin, Q in the body of the tables) Therefore: P(x > Rwf 80) = P(z > 2) = 1-P(z ≤ 2) = 1-0.977 = 0.023

b) The probability that a shopper selected at random spends less than Rwf 50 per week

No need to do any calculations for this question. The mean is Rwf 50 and, because the distribution is normal, so is the median. Half the shoppers, 250, are therefore expected.

Therefore: P(x < Rwf 50) = P(z <0) = 0.500

m = £50, s = £15, x = £80.

5 20 35 50 65 80 95

Q

z

Q

z

z

c) The percentage of shoppers who are expected to spend between Rwf 30 and Rwf 80 per week  Find P(a < Z < b). The probability that a standard normal random variables lies between two values

is also easy to find. The P(a < Z < b) = P(Z < b) - P(Z < a).

For example, suppose we want to know the probability that a z-score will be greater than -1.40 and less than -1.20.

From the Normal probabilistic table, we find that P(Z < -1.20) = 1-P(Z<1.20)= 1-0.8849 = 0.1151 P(Z < -1.20)= 0.1151

and P(Z < -1.40) = 1- P(Z <1.40) = 1-0.9192= 0.0808. P(Z < -1.40) = 0.0808

Therefore, P(-1.40 < Z < -1.20) = P(Z < -1.20) - P(Z < -1.40) = 0.1151 - 0.0808 = 0.0343, therefore the answer is 3.43%

 In our example, we first need P(30 < x < 80) = P(x≤80)-P(x≤30)

Therefore

The whole area is equivalent to100% so 0.8860 of it = 88.6%  The expected number of shoppers who will:

e) Spend less than Rwf 70 per week, First:

Second: Expected number of shoppers who will spend less than Rwf 70 per week: 70*0.9082=63.57

Finding Values from Given Proportions

In this example in parts (a) to (f), we were given the value of x and had to find the area under the normal curve associated with it. Another type of problem gives the area, even if indirectly, and asks for the associated value of x, in this case the value of the shopping basket. Carrying on with the same example we shall find the following information:

 The value below which:

g) 70% of the customers are expected to spend,

The value below which 70% of the shoppers are expected to spend 70% of the total area, 1.00, is below x Þ

From tables, used the other way round:

Approximately to 70% in our normal standard table, correspond the value Z=0.52 (in the margin of the table, by interpolation as z lies between 0.52 and 0.53 and is a little nearer to 0.52)

The value below which 70% of customers spend is 57.80 Rwf The value below which:

h) 350 of the shoppers are expected to spend,

Review problems of chapter

1. What is probability? discuss three applications of probability in everyday life.

2. Give an example of a probability distribution related to something you find interesting, for example, your major, sports, games, etc.

3. The number of computers sold per day at Dan’s Computer Works is defined by the following probability distribution:

x 0 1 2 3 4 5 6

P(x) 0.05 0.10 0.20 0.20 0.20 0.15 0.10

a. , (Answer: p=0.55)

b. , (Answer: p=0.45) c. , (Answer: p=0.75)

d. , (Answer: p=0.55)

4. The table bellow summarizes the community types and degree of pollution for 268 locations in Uganda

Location/pollution Low Moderate High Total

Rural 33 23 9 65

Suburban 8 23 20 51

Urban 7 10 73 90

Commercial 3 11 48 62

Total 51 67 150 268

Suppose that one of the locations is selected randomly. What’s the probability that the locations is a. Suburban? (Answer: p=0.19)

b. Not commercial? (Answer: p=0.77)

c. Rural, given that pollution is low? (Answer: p=0.65)

d. Highly polluted, given that the location is urban? (Answer: p=0.81) e. Urban or highly polluted? (Answer: p=0.62)

f. Rural or not highly polluted? (Answer: p=0.65)

g. Commercial and moderately polluted? (Answer: p=0.04) h. Rural and not highly polluted? (Answer: p=0.21)

5. A marketing manager is considering whether it would be more profitable to distribute his company's product on a national or on a more regional basis. Given the following data, what decision should be made?

Level of Demand

Net profit £m (x)

Prob. that demand is met (p)

E(x)=px _DemandLevel of Net profit_{£m (x)}

Prob. that demand is met (p)

E(x)=px

High 4 0.5 High 2.5 0.5

Medium 2 0.25 Medium 2 0.25

Low 0.5 0.25 Low 1.2 0.25

Expected profits for National distribution =2.625

6. For each of the following, is the random variable discrete or continuous? a. The time required to complete a phone call to a prospective client. b. The number of articles purchased by the client.

c. The shipping weight of the articles. d. The distance the articles are shipped.

7. An automobile dealership in Nairobi has compiled the following sales data over the past year:

Number of cars sold per day

Relative

frequency E(x)

0 0.2 0

1 0.3 0.3

2 0.3 0.6

3 0.15 0.45

4 0.05 0.2

1 1.55

a. What’s the probability there will be more than 2 cars sold in a day? (Answer: p=0.50) b. What is the probability that at least 1 car will be sold in a day? (Answer: p=0.80) c. What is the expected value of the number of cars sold in a day? (Answer: E(x)=1.55)

d. Based on the expected value, how many cars would the dealer expect to sell in a 31 day month? (Answer: 31*1.55=48.05. The dealer should expect to sell about 48 cars a month)

8. The following table comes from the American journal of Economics and Sociology and represents the age distribution of 7581,000 single mothers.

Age Probability Under 18 0.011

18 - 24 0.165 25 - 34 0.402 35 - 44 0.316 45 or more 0.106

1

Find the probability that a single mother is

f. More than 34 years old, and the number of single mothers who are in these intervals? (Answer: p=0.422, n = 3199182)

g. Under 18 or over 45, and the number of single mothers who are in these intervals? (Answer: p=0.117, n = 886977)

9. In order to estimate likely expenditure by customers at a new supermarket, a sample of till slips from a similar supermarket describing the weekly amounts spent by 500 randomly selected customers was analyzed. These data were found to be approximately normally distributed with a mean of Rwf 50 and a standard deviation of Rwf 15. Using this knowledge we can find the following information for shoppers at the new supermarket.

The probability that any shopper selected at random: a) spends more than Rwf 60 per week, (Answer: 0.2514) b) spends less than Rwf 40 per week. (Answer: 0.2514) The percentage of shoppers who are expected to:

c) spend between Rwf 30 and Rwf 60 per week, (Answer: 65.68%) d) spend between Rwf 55 and Rwf 80 per week. (Answer: 34.67%) The expected number of shoppers who will:

e) spend less than Rwf 75 per week, (Answer: 476)

f) spend between Rwf 40.5 and Rwf 60.50 per week. (Answer: 247) The value below which:

g) 60% of the customers are expected to spend. (Answer: 53.75) h) 45% of the customers are expected to spend. (Answer: 48.05) The value expected to be exceeded by:

i) 10% of the customers, (Answer: 30.8) j) 80% of the customers. (Answer: 62.6)

l) 100% of the shoppers are expected to spend (Answer: 108.5)

10.If scores are normally distributed with a mean of 30 and a standard deviation of 5, what percent of the scores is:

(a) greater than 30 (Answer: 50%)? (b) greater than 37 (Answer: 8.08%)? (c) between 28 and 34 (Answer: 44.36%?

11.A Company services copiers. A review of its records shows that the time taken for a service call can be represented by a normal random variable with mean 75 minutes and standard deviation 20 minutes. a. What proportion of service calls taken less than 1 hour? (Answer: 22.66%)

b. What proportion of service call takes more than 90 minutes? (Answer: 22.66%)

c. The probability is 01 that a service call takes more than how many minutes? (Answer: 153 minutes)