Solutions to Additional Questions on Normal Distributions

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Damian Curtis
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1 Solutions to Additional Questions on Normal Distributions 1.. EPA fuel economy estimates for automobile models tested recently predicted a mean of.8 mpg and a standard deviation of mpg for highway driving. Assuming that the mileages are normally distributed (a) what percent of autos should get more than 31 mpg? Soln. Let X = mileage of tested autos. We are given that X N(.8, ). To get the percent of autos that get more than 31 mpg, we need to compute P (X > 31). Standardizing this 31.8 we see that P (X > 31) = P (Z > ) = P (Z > 1) = 1 P (Z < 1), since the area to the right of 1 under the standard normal curve is 1 minus the area to the left of 1 under the standard normal curve. From the Cumulative Normal area table, we get P (Z < 1) = So, P (X > 31) = P (Z > 1) = Hence about % of autos will get more than 31 mpg. (b) what percent of autos should get between 31 and 37 mpg? Soln. Following a similar argument as in (a), we see that we want P (31 < X < 37) = P (1 < Z < 1.97) = P (Z < 1.97) P (Z < 1), since the area between z=1 and z=1.97 is the cumulative area up to 1.97 minus the cumulative area up to 1. From Table 2, we get P (Z < 1.97) = , so that P (31 < X < 37) = = Hence, about 13% of the autos will get between 31 and 37 mpg. (c) describe the gas mileage of the worst 2.5% of all autos. Soln. Call this gas mileage A mpg. Then, we are given that 2.5% of autos have gas mileage under A mpg. [Remember, these are the worst autos, for gas mileage]. Hence, P (X < A) = Translating in terms of the standard normal variable, this is equivalent to P (Z < A.8 ) = From Table 2, the z-score that gives a cumulative area of is So, we have A.8 = 1.96 A = mpg. So, the worst 2.5% of all autos have a gas mileage of mpg or less. 2. Some IQ test scores are known to be normally distributed with a mean of and a standard deviation of. (a) What percent of people should have IQ scores above 1? Soln. Let X = IQ score. Given that X N(, ). We want to find P (X > 1). Standardizing, we get P (X > 1) = P (Z > 1) = as in 1(a). So, about percent of IQ scores will be more than 1. Note that the answer is the same in both cases since we want to find the percent of data that is more than 1 standard deviation from the mean of the data. 1

2 (b) What percent of people should have IQ scores between 68 and 84? Soln. P (68 < X < 84) = P ( 2 < Z < 1) = P (Z < 1) P (Z < 2) = = So, about 14% of people score between 68 and 84. (c) What percent of people should have IQ scores above 132? Soln. P (X > 132) = P (Z > 2) = 1 P (Z < 2) = = So, about 2.3% score above 132. (d) If people above a certain IQ score are chosen for a competition, what should this score be if only 1% of the people are chosen for the competition? Soln. Call this score A. Then, we want this score to be such that P (X > A) = That is, A is the 99th percentile of the IQ scores. This is equivalent to P (Z > A ) = 0.01 P (Z < A ) = From Table 2, the z-score with a cumulative 99 percent of area is So, A = 2.33 A = So, if only 1% is chosen for the competition, then the score cut-off should be The finishing time (in seconds) in men s alpine downhill race at the Salt Lake City Olympics is approximately normally distributed with a mean of seconds and a standard deviation of 3.01 sec. What percent of times is less than 99.7 sec.? Soln. Let X = finishing time (in seconds). Then, X N(102.71, 3.01). So, P (X < 99.7) = P (Z < ) = P (Z < 1) = So, about % of times is less than sec. (4). A company that manufactures rivets believes that the shear strength (in pounds) is modeled as a normal random variable with a mean of 800 lb. and a standard deviation of 50 lb. (a) What percent of rivets would you expect to fall below 900 lb? Soln. Let X = shear strength. (lb.) X N(800, 50).P (X < 900) = P (Z < 2) = So, about 97% of rivets have a shear strength below 900 lb. (b) The strongest 1% of rivets have a shear strength higher than how many pounds? Soln. Call this value A lb. Then, P (X > A) = So, A is again the 99th percentile. Using the same argument as in 2(d), P (X > A) = 0.01 P (Z > A 800 ) = 0.01 P (Z < 50 A 800 ) = 0.99,we find that this value should be A = (2.33) = = lb. So, the strongest 1% rivets have a shear strength of 9.5 lb. or higher. 2

3 5. Assume that the cholesterol levels of adult American women can be described by a normal model with a mean of 188 mg/dl and a standard deviation of mg/dl. (a) What percent of adult women do you expect to have cholesterol level between 150 and 170 mg/dl? Soln. Let X be the cholesterol level. We are given that X N(188, ). So, P (150 < X < 170) = P ( 1.58 < Z < 0.75) = P (Z < 0.75) P (Z < 1.58) = = So, about 17% of adult American women will have cholesterol level between 150 and 170 mg/dl. (b) Find the interquartile range of cholesterol levels. Soln. We want to compute the difference Q 3 Q 1. Q 3 is the third quartile of X and Q 1 is the first quartile of X. By definition, P (X < Q 3 ) = Standardizing, P (Z < Q ) = The z-score with a cumulative area close to 0.75 is So, Q = 0.67 Q 3 = mg/dL. Similarly, by symmetry, Q = 0.67 Q 1 = mg/dL. So, the inter-quartile range is = 32. mg/dl. (c) Above what value are the highest 15 Soln. Call this value A mg/dl. Then, P (X > A) = 0.15 P (Z > A 188 ) = 0.15 P (Z < A 188 ) = The z-score with a cumulative area of approximately 0.85 is So, A 188 = 1.04 A = mg/dL. 6. Suppose the heights of kindergarten children can be described by a normal distribution with a mean of 38.2 inches and a standard deviation of inches. (a) What fraction of kindergarten kids are less than 3 feet tall? Soln. Let X = height of kindergarten children (in inches). X N(38.2, ). Find P (X < 36) z-score of 36 is = So, we get P (X < 36) = P (Z < 1.22) = (b) In what height interval would the middle 80% of kindergarteners heights lie? Soln. We want a value A such that P ( A < X < B) = 0.8. This means that P (X < A) = 0.1 since the area outside the interval [ A, B] is 0.2, half of which is below A. A 38.2 The z-score of A is. The z-score with a cumulative area of approximately A is So, = 1.28 A = A = Similarly, 3

4 P (X < B) = 0.9 P (Z < B 38.2 ) = 0.9 B 38.2 = 1.28 B = So, the middle 80% of the kindergarteners have heights between inches and inches. (c) At least how tall are the biggest 10% of kindergarteners? Soln. Call this height A inches. Then, P (X > X ) = 0.1. From the calculations in part (b), we see that X = inches. So, the biggest 10% of the children are inches tall or taller. (d) 2% of kindergarteners are shorter than what height? Soln. We want a value, say A, such that P (X < A) = P (Z < z ) = The z-score of A is A The z-score with a cumulative area of 0.02 is So, A 38.2 = 2.05 A = 34.51inches. So, 2% of kindergarteners are shorter than inches. 7. Suppose an automobile manufacturer introduces a new model that has an advertised mean city mileage of 27 miles per gallon with a standard deviation of 3 mpg. If you were to buy this model of automobile, what is the probability that you would purchase one that averages less than 20 mpg for city city driving? Soln. Let X = mileage of this model of automobile. We are given that X N(27,3). We want P (X < 20). Standardizing, the z-score of 20 is z = = So, P (X < 20) = 3 P (Z < 2.33). This is the cumulative area up to From the tables, we find this area to be So, the probability of getting a gas mileage of 20 mpg within city with this model is Suppose the scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of. A certain university will consider for admission only those applicants whose score exceed the 90th percentile of the distribution. Find the minimum acceptable score an applicant must achieve in order to receive consideration for admission to the university. Soln. Let X = score of a student on the exam. We are given that X N(550, ). A student is admitted if his or her score is above the 90 th percentile of the scores. Call this 90th percentile as A. By definition, P (X < A) = 0.9. Standardizing, the z-score of A would be A 550 close to 0.9 is So, A 550. So, P (Z < A 550 ) = 0.9. From the Table, the z-score with a cumulative area Solving for A, we get A 678. So, a student has to score at least 678 to gain admission. 9. In studying the dynamics of fish populations, a study published concluded that sardines inhabiting Japanese waters, at two years of age, fish have a length distribution that is approximately normal with a mean length of 20.2 cm. and a standard deviation of

5 cm. Find the probability that a two-year old sardine inhabiting Japanese waters is between 20 and 21 cm. long. Ans A physical-fitness association is including the mile run in its secondary-school fitness test for boys. The time for this event is approximately normally distributed with a mean of 450 seconds and a standard deviation of 40 seconds. If the association wants to designate the fastest 10% as excellent, what time should the association set for this designation? Soln. Let X = time to complete the run. Then, X N(450, 40). Call the time for the excellent designation as A sec. Since this has to represent the fastest 10%, we know that 10% of runners finished A sec. or earlier. So, P (X < A) = 0.1. Standardizing, P ( Z < A 450) = 0.1. From the Table, the z-score with a cumulative area closest to 0.1 is So, A Solving for A, we get A seconds. So, a runner who finishes 40 at or before sec. will be called excellent. 5

The Normal Model. Copyright 2009 Pearson Education, Inc.

The Normal Model. Copyright 2009 Pearson Education, Inc. The Normal Mol Copyright 2009 Pearson Education, Inc. The trick in comparing very different-looking values is to use standard viations as our rulers. The standard viation tells us how the whole collection