1. Up to now, our discussion of the normal distribution has been theoretical. We know how to find the area under the normal bell curve using the normalcdf function. normalcdf(z-left, z-right)

2. We know how to find the z-score that corresponds to the area to the left of a location under the normal bell curve. Invnorm(area to the left)

3. We’ll now look at how the properties of the normal distribution can be applied to real world situations. Real world situations will involve raw scores instead of z-scores. When raw scores are given instead of z-scores, normalcdf changes slightly: normalcdf(lower raw score, upper raw score, mean, standard deviation) E.g. normalcdf(100, 200, 145, 5)

4.  Assume the distribution of tire wear is approximately normal with:  µ = 48,000; σ= 2,000. Find the proportion of tires which would be expected to wear:  (a) Less than 46,000 miles  (b) Greater than 49,000 miles  (c) Between 47,000 and 51,000 miles

5. µ = 48,000; σ= 2,000. (a) Less than 46,000 miles Start by labeling a normal curve with the mean and standard deviation. Locate the raw score in question, and shade the appropriate portion of the curve.

6. Using the picture, list the 4 input values that should be put into normalcdf: Lower raw score: -∞ = -100000=-E99 Upper raw score: 46000 Mean: μ = 48000 Standard deviation: σ = 2000 normalcdf(-E99, 46000, 48000, 2000) = 0.1587

7. µ = 48,000; σ= 2,000. (b) Greater than 49,000 miles Start by labeling a normal curve with the mean and standard deviation. Locate the raw score in question, and shade the appropriate portion of the curve.

8. Using the picture, list the 4 input values that should be put into normalcdf: Lower raw score: = 49000 Upper raw score: ∞ = 100000=E99 Mean: μ = 48000 Standard deviation: σ = 2000 normalcdf(49000, 100000, 48000, 2000) =0.3085

9. µ = 48,000; σ= 2,000. (c) Between 47,000 and 51,000 miles Start by labeling a normal curve with the mean and standard deviation. Locate the raw score(s) in question, (this time, there are two) and shade the appropriate portion of the curve.

10. Using the picture, list the 4 input values that should be put into normalcdf: Lower raw score: = 47000 Upper raw score: = 51000 Mean: μ = 48000 Standard deviation: σ = 2000 normalcdf(47000, 51000, 48000, 2000) = 0.6247

11. A distribution of IQ scores is normally distributed with a mean of 100 and a standard deviation of 15. Find the proportion of IQ scores that are: c) Less than 70 or greater than 145. Identify the mean and standard deviation µ = 100; σ= 15 and label a picture. µ = 100 σ = 15

12. There are two areas that we need to find: Area to the left of 70 and Area to the right of 145 For the area to the left of 70: normalcdf(-E99,70,100,15) =0.0228 µ = 100 σ = 15

13. Area to the right of 145 For the area to the right of 145: normalcdf(145, E99, 100, 15) = 0.0013 Adding the two areas together: = 0.0228 + 0.0013 = 0.0241 µ = 100 σ = 15

14. One thousand students took a standardized psychology exam. The results were approximately normal with µ = 83 and σ= 8. Find the number of students who scored: a) Less than 67 b) Greater than 87

15. a) Less than 67 Label the diagram with the raw score in question. Shade the appropriate area. Area is found with: normalcdf(-E99,67,83,8) = 0.0228 We need to find the number of students, not the percentage. µ = 83 σ = 8

16. a) Less than 67 The number of students less than 67 is: 1000 * 0.0228 = 22.8 = 23 students µ = 83 σ = 8

17. b) Greater than 87 Label the diagram with the raw score in question. Shade the appropriate area. Area is found with: normalcdf(87,E99,83,8) = 0.3086 We need to find the number of students, not the percentage.

18. b) Greater than 87 The number of students greater than 87 is: 1000 * 0.3086 = 308.6 = 309 students

19. Recall from Chapter 3 that the percentile rank of the data value X indicates the percent of data value within the distribution less than the data value X. In a normal distribution, the percentile rank of a data value X is equal to the percent of area under the normal curve to the left of (below) the data value X. Percentile ranks are always expressed as a whole number. X Raw scores Area to the left of X is the percentile rank of X

20. Matthew earned a grade of 87 on his history exam. If the grade in his class were normally distributed with µ = 80 and σ= 7, find Matthew’s percentile rank on this exam. The percentile rank of a data value X=87 is equal to the percent of area under the normal curve to the left of (below) the data value X=87.

21. The percent of area under the normal curve to the left of (below) 87: normalcdf(-E99,87,80,7) = 0.8413 Convert to a percent: 84.13% Round to the nearest whole number : 84 So, the percentile rank of 87 is 84. In other words, the score 87 is the 84 th percentile. In percentile notation, 87=P 84

22. Miss Brooke took two exams last week. Her grades and class results were as follows: Assuming the results of both exams to be normally distributed, use percentile ranks to determine on which exam Miss Brooke did better relative to her class.

23. To find the percentile rank of the English exam, calculate the area to the left of the raw score 83. The percent of area under the normal curve to the left of (below) 83 is: normalcdf(-E99,83,80,6) 0.6915 = 69.15% or a percentile rank of 69

24. To find the percentile rank of the Math exam, calculate the area to the left of the raw score 77. The percent of area under the normal curve to the left of (below) 77 is: normalcdf(-E99,77,74,5) 0.7257 = 72.57% or a percentile rank of 73

25. The percentile rank of 69 for the English grade means that 69% of the class scored below Miss Brookes. The percentile rank of 73 for the Math grade means that 73% of the class scored below Miss Brookes. So which grade is better, relative to the class? The math grade with with the higher percentile rank is better.

26. Consider a normal distribution with µ = 75 and σ= 5. Find the data value in the distribution that: (a) Cuts off the top 25% of the data values. (b) Represents the 30 th percentile. (c) Has a percentile rank of 55. (d) Cuts off the middle 40% of the data values.

27. (a) Cuts off the top 25% of the data values. The raw score that cuts off the top 25% is the same as the raw score that cuts off the bottom 75%. Label a diagram: We are asked for a raw score and we have the area to the left: 0.75 Use invnorm to get the raw score that cuts off the bottom 75% Bottom 75%

28. (a) Cuts off the top 25% of the data values. When using raw scores instead of z-scores, invnorm changes slightly: Invnorm(area to the left, mean, standard deviation) = invnorm(0.75, 75, 5) = 78.37

29. (b) Represents the 30 th percentile. Recall that the 30 th percentile is the data value that has 30% of the distribution below it. Label a diagram: We are asked for a raw score and we have the area to the left: 0.30 We use invnorm to get the raw score that cuts off the bottom 30%

30. (b) Represents the 30 th percentile. = invnorm(0.30, 75, 5) = 72.38

31. (c) Has a percentile rank of 55. The data value that has a percentile rank of 55 cuts off the bottom 55% of the data values in the distribution. Label a diagram: We are asked for a raw score and we have the area to the left: 0.55 We use invnorm to get the raw score that cuts off the bottom 55%

32. (c) Has a percentile rank of 55. = invnorm(0.55, 75, 5) = 75.63