1. Chapter 5 Normal Probability Distributions

2. Chapter 5 Normal Probability Distributions Section 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution A.The normal distribution is the most important of the continuous probability distributions. 1.Definition: A normal distribution is a continuous probability distribution for a random variable x. a.The graph of a normal distribution is called a normal curve. 2.A normal distribution has the following properties: a.The mean, median, and mode are equal (or VERY close to equal). b.The normal curve is bell-shaped and is symmetric about the mean. c.The total area under the normal curve is equal to one. d.The normal curve approaches, but never touches, the x-axis as it gets further away from the mean.

3. Chapter 5 Normal Probability Distributions Section 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution e.The graph curves downward within one standard deviation of the mean, and it curves upward outside of one standard deviation from the mean. 1)The points where the curve changes from curving upward to curving downward are called inflection points. B.We know that a discrete probability can be graphed with a histogram (although we didn’t emphasize this in Chapter 4). 1.For a continuous probability distribution, you can use a probability density function (pdf). a.A probability density function has two requirements: 1) The total area under the curve has to equal one. 2)The function can never be negative.

6. Chapter 5 Normal Probability Distributions Section 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution 2.The standard normal distribution has the following properties: a.The cumulative area under the curve is close to 0 for z-scores close to -3.49. b.The cumulative area increases as the z- scores increase. c.The cumulative area for z = 0 is 0.5000. d.The cumulative area is close to 1 for z-scores close to 3.49. 3.To find the corresponding area under the curve for any given (or calculated) z-score, there are two main methods. a.The easiest, and the one I suggest, is to use the TI-84 calculator. 1)2 nd VARS normalcdf (lower boundary, upper boundary) a)If you want the area to the left of a z-score, use -10,000 as your lower boundary and the z-score you are interested in as your upper boundary.

7. Chapter 5 Normal Probability Distributions Section 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution b)If you want the area to the right of a z-score, use z- score you are interested in as your lower boundary and use 10,000 as your upper boundary. c)If you want the area between two z-scores, use them both (smaller one as lower, larger one as upper). b.The other way, which will be demonstrated in class, is to use the z-score chart. 1)You should have done this in Algebra 2, so it should be a quick review for us.

8. Chapter 5 Normal Probability Distributions Section 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution 4.Remember that in Section 2.4 we learned from the Empirical Rule that values lying more than 2 standard deviations from the mean are considered to be unusual. a.We also learned that values lying more than 3 standard deviations from mean are very unusual, or outliers. b.In terms of z-scores, this means that a z-score of less than -2 or greater than 2 means an unusual event. 1)A z-score of less than -3 or greater than 3 means a very unusual event. (Outlier)

9. Section 5-1 Normal Probability Distributions Examples

10. Page 249 Selected Even Problems, Using the TI-83 #22 Find the area to the left of z = 0.08. 2 nd VARS normalcdf(-10000,.08) =.5319 #32 Find the area to the right of z = 2.51 2 nd VARS normalcdf(2.51,10000) =.0060 #36 Find the area between z = -0.51 and z = 0 2 nd VARS normalcdf(-0.51,0) =.1950 #40 Find the area to the left of z = -1.96 or to the right of z = 1.96 Remember that we ADD probabilities for OR questions. 2 nd VARS normalcdf(-10000,-1.96) =.025 2 nd VARS normalcdf(1.96,10,000) =.025.025 +.025 =.0500

11. You are performing a study about the height of 20-29 year old men. A previous study found the height to be normally distributed, with a mean of 69.6 inches and a standard deviation of 3.0 inches. You randomly sample 30 men and find their heights (in inches) to be as follows: 72.171.267.967.369.568.668.869.473.567.1 69.275.771.169.670.766.971.462.969.264.9 68.265.269.772.267.566.666.564.265.470.0 A)Draw a frequency histogram to display these data points using seven classes. Is it reasonable to assume that the heights are normally distributed? Why? B)Find the mean and standard deviation of your sample. C)Compare the mean and standard deviation of your sample with those in the previous study. Discuss the differences. Page 250, # 42

12. You are performing a study about the height of 20-29 year old men. A previous study found the height to be normally distributed, with a mean of 69.6 inches and a standard deviation of 3.0 inches. You randomly sample 30 men and find their heights (in inches) to be as follows: 72.171.267.967.369.568.668.869.473.567.1 69.275.771.169.670.766.971.462.969.264.9 68.265.269.772.267.566.666.564.265.470.0 Page 250, # 42 Entering the 30 data points into the TI-83, using STAT and Edit, we can calculate the mean, standard deviation, and median. STAT, Calc, 1-Var Stats gives us what we need. The mean is 68.75, the standard deviation is 2.85, and the median is 69.00.

13. LLULLBUBMdPtFreq.Rel. Freq. Cum. Freq. Max Value: 75.7 Min Value: 62.9 Range: 75.7 – 62.9 = 12.8 Class Width: 12.8/7 = 2 Remember to ROUND UP!! First Lower Limit is the Minimum Value!!! Add Class Width Down 64.9 66.9 68.9 70.9 72.9 62.9 64.8 First Upper Limit is one unit less than the 2 nd Lower Limit (Remember, our units are tenths, not whole numbers). Add Class Width Down 66.8 68.8 70.8 72.8 74.8 72.171.267.967.369.568.668.869.473.567.1 69.275.771.169.670.766.971.462.969.264.9 68.265.269.772.267.566.666.564.265.470.0 74.976.8

14. LLULLBUBMdPtFreq.Rel. Freq. Cum. Freq. 64.9 66.9 68.9 70.9 72.9 62.9 64.8 66.8 68.8 70.8 72.8 74.8 72.171.267.967.369.568.668.869.473.567.1 69.275.771.169.670.766.971.462.969.264.9 68.265.269.772.267.566.666.564.265.470.0 74.976.8 Subtract one-half unit from lower limits to get lower boundaries. REMEMBER that our units are tenths!! One-half of a tenth is 5 hundredths (.05) Add one-half unit to upper limits to get upper boundaries 62.85 64.85 66.85 68.85 70.85 72.85 74.85 64.85 66.85 68.85 70.85 72.85 74.85 76.85 Find the mean of the limits (or boundaries) to find the midpoint of each class. 63.85 65.85 67.85 69.85 71.85 73.85 75.85

15. Count how many data points fit in each class and enter that into the Frequency column LLULLBUBMdPtFreq. 64.9 66.9 68.9 70.9 72.9 62.9 64.8 66.8 68.8 70.8 72.8 74.8 74.976.8 62.85 64.85 66.85 68.85 70.85 72.85 74.85 64.85 66.85 68.85 70.85 72.85 74.85 76.85 63.85 65.85 67.85 69.85 71.85 73.85 75.85 72.171.267.967.369.568.668.869.473.567.1 69.275.771.169.670.766.971.462.969.264.9 68.265.269.772.267.566.666.564.265.470.0 2 5 8 8 5 1 1

16. 88 55 2 11 Looking at the histogram drawn from the frequency table, it is easy to see that the data is almost perfectly bell- shaped, symmetrical and centered about the mean. Draw the histogram using the frequencies obtained from the table we just did. The mean, median, and mode are also very closely bunched together. Mean is 68.75, median is 69.00 and the mode is 69.2 For these reasons, it is reasonable to assume that the heights are normally distributed. 62.8564.8566.8568.8570.8572.8574.8576.85

17. The last part of the question was to compare the mean and standard deviation of your sample with those in the previous study. Discuss the differences. Our mean and standard deviation were 68.75 and 2.85. The previous study had a mean of 69.6 and a standard deviation of 3.0. This means that our sample of men was shorter than the previous study, but that they were also more closely bunched together in height.

18. Your assignments are: Classwork: Pages 248-250, #1-8 All, and #9-41 Odd Homework: Pages 250-252, #43-61 Odd