Let X be a normal distribution with mean µ and standard deviation s. For a given probability p, we often wish to find values a, b, c, and d such that P(X <= a) = p, or P(b <= X) = p, or P(c <= X <= d) = p. The value a is called the p quantile. If we convert p to a percentage, then a is called the 100*p percentile. For example, the 0.90 quantile (or the 90th percentile) is the value a such that P(X <= a) = 0.90.
We can find any of these values a, b, c, or d using the INVNORM program that computes three types of inverse normal calculations. To execute the program, first enter 1, 2, or 3 to specify that you want the inverse from a left probability, or the inverse from a right probability, or the inverses from a middle probability.
If you enter 1, then the program will find the value a such that P(X <= a) = p. If you enter 2, then the program will find the value b such that P(b <= X) = p. If you enter 3, then the program will find the values c and d such that P(c <= X <= d) = p.
After specifying either 1, 2, or 3, enter the desired probability p, followed by the mean and the standard deviation of the normal distribution.
Example. Heights are generally found to be normally distributed. Assume that the average adult female height is 65.5 inches with a standard deviation of 3 inches.
(a) What height do 90% of women fall below?
(b) What height do 60% of women measure above?
(c) What two heights, symmetric about the mean, contain 80% of all women's heights?
Solution. (a) We can find either the left inverse probability value for p = 0.90 or the right inverse probability value for p = 0.10. After calling up the INVNORM program, enter 1 to specify a left probability, then enter .9 for this left probability. Next enter 65.5 for MEAN and 3 for STANDARD DEV. We receive an inverse value of about 69.3447. Hence, 90% of adult females should be below 69.3447 inches tall.
(b) We can find either the right inverse probability value for p = 0.60 or the left inverse probability value for p = 0.40. First enter 2 to specify a right probability, then enter .6 for this right probability. Next enter 65.5 for MEAN and 3 for STANDARD DEV. We receive an inverse value of about 64.74. Hence, 60% of adult females should measure above 64.74 inches tall.
(c) First enter 3 to specify a middle probability, then enter .8 for this middle probability. Then enter the mean and standard deviation. We receive inverse values of 61.655 and 69.345. Hence, 80% of adult females should measure between 61.655 and 69.345 inches tall.
1. Suppose adult males average 68.5 inches tall with a standard deviation of 3.25 inches.
(a) What is the height for which 75% of men measure at least this height?
(b) What two heights, symmetric about the mean, contain 50% of all men's heights?
2. Suppose the mean SAT score during a given year was 980 with a standard deviation of 190. As usual, the scores were normally distributed.
(a) Find the cutoff for the 80th percentile.
(b) Find the two scores that are symmetric about the mean that contain 96% of all scores. What percentile is this upper score?
3. Packaged weights of Cheer detergent are normally distributed with a mean of 42 oz. and a standard deviation of 0.80 oz. But the marked weight on the package is such that only about 5.9% of packages contain less than this weight. What is the marked weight?
1. (a) 66.308 inches
(b) 66.308 in. to 70.692 in.
2. (a) 1140 (computed as1139.908)
(b) Roughly 590 to 1370 (the 98th percentile).
3. 40.75 oz.
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