1. 16-1 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Chapter 16 The Normal Distribution Introductory Mathematics & Statistics for Business

2. 16-2 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Learning Objectives Identify the properties of the normal distribution and normal curve Identify the characteristics of the standard normal curve Understand examples of normally distributed data Read z-score tables and find areas under the normal curve Find the z-score, given the area under the normal curve Compute proportions Check whether data follow a normal distribution Understand and apply the Central Limit Theorem Solve business problems that can be represented by a normal distribution Calculate estimates and their standard errors Calculate confidence intervals for the population mean Calculate confidence intervals for the population proportion

3. 16-3 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.1 Introduction When the frequencies of observations for a large population result in a frequency polygon that follows the pattern of a smooth bell-shaped curve that population is said to have a normal distribution.

4. 16-4 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Introduction (cont…) The normal curve –If the frequency polygon for a given set of observations that have a normal distribution is made into a smooth curve, the resulting curve is referred to as a normal curve. –A specific normal curve is characterised by its mean (μ) and standard deviation (σ).

5. 16-5 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Introduction (cont…) The main features of a normal distribution are: –It is bell-shaped. –It is symmetric about the mean. –It is asymptotic to the horizontal axis. –Approximately 68% of the distribution lies within 1 standard deviation of the mean; –about 95% lies within 2 standard deviations of the mean; –about 99.7% lies within 3 standard deviations of the mean. –The location and dispersion will depend on the values of μ and σ (see Figures 16.2 and 16.3). – The total area under any normal curve is 1.

6. 16-6 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.2 An Example of Data that have an Approximate Normal Distribution As part of a study by the author, random samples of 200 males and 200 females were selected from the students enrolled in a large introductory statistics course at Macquarie University, Sydney. Each student was asked to record his or her height, to the nearest centimetre. The male heights have a mean of 178.6 cm and a standard deviation of 7.0 cm, and the female heights have a mean of 164.5 cm and a standard deviation of 7.7 cm.

7. 16-7 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e An Example of Data that have an Approximate Normal Distribution (cont…) The histograms displaying the grouped frequency distributions for the male and female heights

8. 16-8 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.3 Areas Under The Normal Curve The proportions of observations that take on certain values are represented by areas under the distribution curve The proportion of observations that take on a value between a and b is the area under the curve between two vertical lines erected at a and b. We could calculate these areas and thus obtain values for the proportions.

9. 16-9 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Areas Under The Normal Curve (cont…) Standard Scores (z – scores) –The z-score of a measurement is defined as the number of standard deviations the measurement is away from the mean –If the measurement is above the mean, the corresponding z- score is positive; but if the measurement is below the mean, the corresponding z-score is negative –Thus, if a distribution has a mean of μ and a standard deviation of σ, the corresponding z-score of an observation x is:

10. 16-10 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Areas Under The Normal Curve (cont…) Standard Scores (z – scores) –The mean itself has a z-score of zero and that a value exactly 1 standard deviation from the mean has z = + 1 or – 1. –The mean itself has a z-score of zero and that a value exactly 1 standard deviation from the mean has z = + 1 or – 1. The larger a positive z-value, the further the given x is above the mean, while large negative z-values correspond to extreme values below the mean –It is found in practice with most data sets that the vast majority of observations have a z-score in the range roughly –2 to + 2.

11. 16-11 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.4 Using the Table of the Standard Normal Curve Table 6 gives the area under the standard normal curve between the mean (zero) and any positive number z. –Example Using Table 6, determine the area under the standard normal curve: (a) between z = 0 and z = 1.50 (b) between z = –2.10 and z = 0 (c) between z = 0.60 and z = 1.80 (d) between z = –0.30 and z = 2.25 (e) to the right of z = 1.95

12. 16-12 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Using the Table of the Standard Normal Curve (cont…) Solution (a) From Table 6, the area between z = 0 and z = 1.50 is 0.4332 (b) In order to find the area to the left of z = 0, the symmetry of the normal curve can be used. In this case, the area between z = –2.10 and z = 0 is the same as the area between z = 0 and z = + 2.10. From Table 6, this area is 0.4821. (c) The area between z = 0.60 and z = 1.80 may be found by subtracting the area between z = 0 and z = 0.60 from the area between z = 0 and z = 1.80. From Table 6: The area between z = 0 and z = 1.80 is 0.4641. The area between z = 0 and z = 0.60 is 0.2257. Therefore, the required area is 0.4641 – 0.2257 = 0.2384.

13. 16-13 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Using the Table of the Standard Normal Curve (cont…) Solution (cont…) (d) The area between z = –0.30 and z = 2.25 may be found by adding the area between z = – 0.30 and z = 0 to the area between z = 0 and z = 2.25. The area between z = – 0.30 and z = 0 equals the area between z = 0 and z = + 0.30 (by symmetry). From Table 6: The area between z = 0 and z = 0.30 is 0.1179. The area between z = 0 and z = 2.25 is 0.4878. Therefore, the required area is 0.1179 + 0.4878 = 0.6057. (e) From the facts that the normal curve is symmetrical about the mean and the total area under the curve is 1, it follows that the area to the right of z = 0 is 0.5 and that the area to the left of z = 0 is also 0.5. Thus, to find the area to the right of z = 1.95, we subtract the area between z = 0 and z = 1.95 from 0.5. From Table 6: The area between z = 0 and z = 1.95 is 0.4744. Therefore, the required area is 0.5000 – 0.4744 = 0.0256.

14. 16-14 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Using the Table of the Standard Normal Curve (cont…) Conversion to raw scores –In order to determine appropriate areas under any normal curve, the z-score (or standard score) may be calculated. –The z-scores express the given problem in ‘standard form’ so that the standard normal curve can be used.  To convert a raw score of x (from a distribution with mean μ and standard deviation σ) to a z-score, subtract the mean from x and divide by the standard deviation.  To convert a z-score to a raw score x, multiply the z- score by the standard deviation and add this product to the mean. –In equation form:

15. 16-15 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.5 Computation of Proportions A proportion within a given interval –If a set of data has a normal distribution, the proportion of observations that lie in a particular interval can be found using the following procedure: 1.Determine the z-score for each end point of the interval. 2. Find the area (from Table 6) for each z-value. (If the z- value is negative, ignore the sign.) 3. lf the end points of the interval lie on opposite sides of the mean, add the two areas found in Step 2. If the two end points lie on the same side of the mean, subtract the smaller area from the larger one.

16. 16-16 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Computation of Proportions (cont…) A proportion greater or less than a given value –Table 6 can also be used to determine what proportion of a normal population is greater than a certain value or less than a certain value  To determine the proportion of a normal distribution greater than a value of x, calculate the z-score corresponding to x and find the area to the right of this score.  To determine the proportion of a normal distribution less than a value of x, calculate the z-score corresponding to x and find the area to the left of this score.

17. 16-17 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Computation of Proportions (cont…) A value corresponding to a given proportion –For some problems it is necessary to find the z-score for a given area rather than the other way around.  Example : Suppose that a particular MBA entrance exam is designed so that the marks will be normally distributed with a mean of 300 and a standard deviation of 60. The policy is that the top 35% gain admission to the program. What is the lowest mark that would gain admission?  Solution  From Table 6, z = 0.38 gives an area of 0.1480 while z = 0.39 gives an area of 0.1517. Thus, the z-score required is somewhere between 0.38 and 0.39, say 0.385. this z-score can be converted to a raw score:

18. 16-18 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.6 Checking Whether Data are Normally Distributed Even if the distribution is believed to be normal, it is rare for the mean μ and standard deviation s to be known precisely much of statistical testing depends on whether or not the data are normally distributed A wrong assumption can lead to an incorrect statistical test being used on the data, which in turn can lead to a doubtful conclusion The construction of a histogram and corresponding frequency polygon of the data can give a rough idea of its shape If the frequency polygon for a large sample clearly does not have a normal shape, it is unlikely that the population does either.

19. 16-19 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.7 The Central Limit Theorem Definition –If random samples of size n are selected from a population with mean μ and standard deviation σ, the means of the samples are approximately normally distributed with mean μ and standard deviation even if the population itself is not normally distributed, provided that n is not too small. The approximation becomes more and more accurate as the sample size n is increased. –The Central Limit Theorem also says that any unusual shapes of populations are suppressed. That is, even if individual values come from a very non-normal population (such as one that is skewed or bimodal), the sample means tend to have a normal distribution.

20. 16-20 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e The Central Limit Theorem (cont…) Conversion of a sample mean to a z-score –The formula for converting a mean of to a z-score is given by: –Where

21. 16-21 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.8 Confidence Intervals for an Unknown Population Mean Point estimates –A single estimate of an unknown population mean can be obtained from a random sample –Different random samples give different values of the mean –A single estimate is referred to as a point estimate –Accuracy depends on:  variability of data in the population  size of the random sample

22. 16-22 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Mean (cont…) The Standard error of the mean Standard error of the mean provides the precise measure of accuracy of a point estimate of the mean the value of σ can be replaced by the sample standard deviation, s. where: σ = population standard deviation n = size of random sample

23. 16-23 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Mean (cont…) The meaning of confidence intervals for μ –Instead of providing a single point estimate for μ, we consider a range of values or an interval within which the value of μ may lie. –Be able to provide a probability or level of confidence that this interval does indeed contain the true value of μ. –Common probabilities to use are 0.90, 0.95 and 0.99 –Confidence intervals for μ are based on the values from random samples

24. 16-24 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Mean (cont…) Construction of confidence intervals for μ –Suppose that we have the following information:  The population has a normal distribution.  The population standard deviation (σ) is known.  The sample size (n) can be of any size. –Then a 95% confidence interval for μ is: x = the mean of the sample σ = the population standard deviation n = the size of the sample

25. 16-25 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 16.9 Confidence Intervals for an Unknown Population Proportion Point estimates –To give a single estimate of an unknown population proportion (π), use the value of the proportion (p) obtained from a random sample taken from that population. –A single estimate of π is referred to as a point estimate. –Point estimates are particularly important in survey work, for example, to get an idea of how the population sampled feels about a certain issue. –How accurate a point estimate is depends on two factors:  how variable the data are in the population  the size of the random sample used to make the estimate

26. 16-26 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Proportion (cont…) The standard error of the proportion –The precise measure of accuracy of a point estimate (p) of a population proportion is provided by the standard error of the proportion –The formal definition is –Where π = the population proportion n = the size of the random sample the value of π can be replaced by the sample standard proportion, p

27. 16-27 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Proportion (cont…) The meaning of confidence intervals for π –Instead of providing a single point estimate for π, we consider a range of values or an interval within which the value of π may lie. –Be able to provide a probability or level of confidence that this interval does indeed contain the true value of π. –Common probabilities to use are 0.90, 0.95 and 0.99, although any probability could be used. –Confidence intervals for π are based on the values from random samples

28. 16-28 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Confidence Intervals for an Unknown Population Proportion (cont…) Construction of confidence intervals for π –We are able to construct an interval estimate for the true value of an unknown population proportion π. –The confidence interval for π is of the form –where the value of z is chosen as follows:  z = 1.645 for a 90% confidence interval  z = 1.96 for a 95% confidence interval  z = 2.58 for a 99% confidence interval

29. 16-29 Copyright  2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e Conclusion Look at identifying the properties of the normal distribution and normal curve Also identified the characteristics of the standard normal curve Understood examples of normally distributed data Read z-score tables and find areas under the normal curve Found the z-score, given the area under the normal curve Computed proportions Checked whether data follow a normal distribution Understood and applied the Central Limit Theorem Solved business problems that can be represented by a normal distribution Calculated estimates and their standard errors Calculated confidence intervals for the population mean Calculated confidence intervals for the population proportion