TIME FRAME: 60 minutes
LEARNING COMPETENCIES
At the end of the lesson, learners should be able to:
• compute probabilities using a table of cumulative areas under a standard normal curve
• compute percentiles of a (standard) normal curve
PRE-REQUISITE LESSONS: Random Variables, Probability Distribution of Continuous Random Variables, Properties of Normal Distributions
LESSON OUTLINE
A. Introduction: Review of Normal Distribution B. Main Lesson: Areas Under a Normal Curve C. Enrichment: Computing with Excel
REFERENCES
Albert, J. R. G. (2008). Basic Statistics for the Tertiary Level (ed. Roberto Padua, Welfredo Patungan, Nelia Marquez). Philippines: Rex Bookstore.
De Veau, R. D., Velleman, P. F., and Bock, D. E. (2006). Intro Stats. Pearson Ed. Inc.
Workbooks in Statistics 1: 11th Edition, Institute of Statistics, UP Los Baños, College Laguna 4031
Probability and statistics: Module 22. (2013). Australian Mathematical Sciences Institute and Education Services Australia. Retrieved from
http://www.amsi.org.au/ESA_Senior_Years/PDF/ExpoNormDist4f.pdf
DEVELOPMENT OF THE LESSON
A. Introduction: Review of Normal Distribution
Ask learners to recall some of the lessons learned about normal distributions. They should be able to state that:
• A normal distribution has a symmetric bell-shaped curve (for its probability density function) with one peak. This curve is characterized by its mean m (the center of symmetry, and also the peak) and standard deviation σ (the distance from the center to the change-of-curvature points on either side). If a random variable X has a normal distribution with mean m and variance σ2, we denote this as X~N(μ,σ2).
• A normal curve is symmetric about its mean (thus the mean is the median). It is more concentrated in the middle and its peak is at the mean (so that the mean is also the mode).
• Like any continuous distribution, the total area under the normal curve is equal to 1, and the probability that a normal random variable X equals any particular value a , P(X=a) is zero (0).
• The normal curve follows the empirical rule (also called the 68-95-99.7 rule):
o About 68% of the area under the curve falls within 1 standard deviation of the mean.
o About 95% of the area under the curve falls within 2 standard deviations of the mean.
o Nearly the entire distribution or about 99.7% of the area under the curve falls within 3 standard deviations of the mean.
B. Main Lesson: Probabilities/Areas Under a Normal Curve
Inform learners that there are countless normal curves, and we could very readily obtain the probabilities with computers, specifically with the use of spreadsheet applications of statistical software applications like Microsoft Excel.
They need to examine a special normal distribution called the standard normal distribution.
The Standard Normal Curve
Define the standard normal distribution to be the normal distribution with a mean of 0 and a standard deviation of 1, and draw a standard normal curve:
Tell learners that the notation Z ~N (0; 1) means that the random variable Z has a standard normal distribution, i.e. m = 0 and s = 1. Ask learners to use the empirical rule to determine the following areas under a standard normal curve:
(a) -1 to +1 (Answer should be 68 percent) (b) - 2 to +2 (Answer should be 95 percent) (c) - 3 to +3 (Answer should be 99.7 percent)
Provide a copy of the handout (found at the end of this lesson plan) to learners.
This is a table of the cumulative distribution function (i.e. the area to the left of some particular value z) of a standard normal curve. That is, this table reports the cumulative probability associated with a particular z-score:
Φ(z) = P(Z ≤ _z) = area under a Standard normal curve to the left of some particular z
Explain to learners that the table’s rows show the whole number and tenths place of the z-score, while the table’s columns show the hundredths place, and finally, the cumulative probability Φ(z) appears in the cell of the table.
For example, a section of the standard normal table is reproduced below. To find the cumulative probability of a z-score equal to -1.31, explain to students that they should cross-reference the row of the table containing -1.3 with the column
containing 0.01. The table shows that the probability that a standard normal random variable will be less than -1.31 is 0.0951; that is, Φ(1.31) = P(Z ≤ -1.31) = 0.0951.
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823
Practice further with learners on how to use this table of cumulative probabilities under a standard normal curve. Assume that we have a random variable Z that has a standard normal distribution. Ask them what would be:
(a) P( Z ≤ 0 ): Answer should be 0.5 since the first entry of the first line (of the second page) for the Table of values of Φ(z) reads so.
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
(b) P( Z ≤ -1.54 ) ; As per Table of values of Φ(z), answer is 0.0618
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
-1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559
(c) P(-1.54 ≤ Z ≤ 1.54 ) = 0.8764. Get a graph of the pertinent area of interest, and show that the area between -1.54 and 1.54 can be obtained from the difference of the area to the left of 1.54 and the area to the left of -1.54:
= P( Z ≤ 1.54 ) - P( Z ≤ -1.54 ) = 0.9382- 0.0618 (as per the table entries) = 0.8764
(d) P(Z ≥ 1.54) = 0.0618
P(Z ≥ 1.54) is an upper tail area, but the total area under the curve is 1, so P( Z ≥ 1.54 ) is the difference of 1 and the area to the left of 1.54, i.e.
1- P( Z ≤ 1.54 ) = 1 - 0.9382.= 0.0618
Technical Computing Note:
Percentiles may also be obtained from the table. For instance, illustrate this by giving an example such as this: obtaining values of z for which (a) the area to the left of z in a standard normal curve is 0.5832; (b) the area to the right of z is 0.8508.
Show that we can find z directly for (a), by looking for the value of z that gives an entry of 0.5832. In this case, we find z to be 0.21.
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
For (b), show firstly that if the area to the right of z is 0.8508, then the area to the left of z is 0.1492, so that consequently they need to observe that z=-1.04.
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
-1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379
Computing Note:
If computers are available, show learners that we could alternatively use Excel to obtain (a) and (b). Merely enter the command
= NORMSINV(0.5832)
and generate the value of z as 0.210086 for (a).
While for (b), we enter the command
= NORMSINV(1-0.8508) and thus find z as –1.03987.