In-class Exercises on Normal Distributions
October 29, 2007
Name. Hints and Answers
1. Assume the distance a golfer drives a golf ball is normally distributed with a mean of 850
feet and a standard deviation of 25 feet.
900 − 850
=2
(a) Find the z-score for a drive of 900 feet. z =
25
825 − 850
(b) Find the z-score for a drive of 825 feet. z =
= −1
25
820 − 850
(c) Find the z-score for a drive of 820 feet. z =
= −1.2
25
(d) If the golfer had a drive that was 2.5 standard deviations above the mean, how long was
the drive (note z = 2.5 in this case)?
x = 850 + 2.5(25) = 912.5ft.
(e) If the golfer had a drive that was 1.5 standard deviations below the mean, what is the
z-score and how long was the drive?
x = 850 − 1.5(25) = 812.5ft.
2. Use the table for the normal distribution to compute the following probabilities.
(a) P (z ≤ 1.24) = .8925
(b) P (0 ≤ z ≤ 1.6) = .9452 − .5000 = .4452
(c) P (−2.37 ≤ z ≤ 0) = .5000 − .0089 = .4911
(d) P (−3 ≤ z ≤ 3) = .9987 − .0013 = .9974
(e) P (−2.34 ≤ z ≤ 1.57) = .9418 − .0096 = .9322
(f) P (1.24 ≤ z ≤ 1.88) = .9699 − .8925 = .0774
(g) P (−2.44 ≤ z ≤ −.73) = .2327 − .0073 = .2254
(h) P (z ≤ 1.64) = .9495
(i) P (z ≥ 2.39) = 1 − .9916 = .0084
(j) P (z ≥ −1.43) = 1 − .0764 = .9236
(k) P (z < −2.71) = .0034
3. (Data from p. 331# 23) The SAT mathematics test has a mean score of 500 with a standard
deviation of 100. Assume that the scores are normally distributed, and then find
(a) The probability a randomly selected SAT math score will be between 450 and 636.
(b) The probability a randomly selected SAT math score will be more than 750.
(c) The probability a randomly selected SAT math score will be less than 636.
Answer. For this problem, µ = 500 and σ = 100.
(a) If x = 450, z =
636 − 500
450 − 500
= −.50 and if x = 636, z =
− 1.36. Therefore,
100
100
P (450 < x < 636) = P (−.5 < z < 1.36) = .9131 − .3085 = .6046
(b) If x = 750, then z =
750 − 500
= 2.5. Therefore,
100
P (x > 750) = P (z > 2.5) = 1 − .9938 = .0062
(c) P (x < 636) = P (z < 1.36) = .9131
4. Use the table for the standard normal distribution to answer the following.
(a) Find z so that 6% of the standard normal curve lies to the left of z.
(b) Find z so that 6% of the standard normal curve lies to the right of z.
(c) Find z so that 98% of the standard normal curve lies between −z and z.
(d) Find z so that 45% of the standard normal curve lies between 0 and z.
(e) Find z so that 95% of the standard normal curve lies to the left of z.
(f) Find z so that 95% of the standard normal curve lies to the right of z.
Answer. (a) Find z corresponding to an area of .06, so z ≈ −1.555.
(b) Find z corresponding to an area of .94 to left of z, so z ≈ 1.555.
(c) If 98% is between −z and z, then 1% is to right of z and 99% to the left of z, and so
z ≈ 2.33.
(d) If 45% is between 0 and z, then 95% is to the left of z, and so z ≈ 1.645.
(e) Same as (d), z ≈ 1.645.
(f) z ≈ −1.645, the negative of (e).
5. (Data from Section 6.3 # 24) Suppose the mathematics SAT scores are normally distributed with a mean of 500 and a standard deviation of 100, and the ACT scores are normally
distributed with a mean of 18 and a standard deviation of 6.
(a) If a school only accepts students with SAT or ACT math scores in the top 10 percent, what
are the minimum ACT and SAT scores that would be acceptable for that school.
(b) Same question, but for a school accepting the top 20%.
(c) Same question, but for a school accepting the top 60%.
Hint. For (a), find z to that 90% of standard normal curve lies to the left of z. Thus, z = 1.28,
and so
ACT score must be at least 18 + 1.28(6) = 25.68 or 26.
SAT score must be at least 500 + 1.28(100) = 628.
(b) Use same method, but this time z = .84, then x = µ + .84σ for each test. Thus the
minimum SAT score is 584 and the minimum ACT score is 23.04.
(c) Use same method, but this time z = −.25 (this is the z-value closest to the 40th percentile),
and then x = µ − .25σ for each test. The minimums are then 475 on the SAT, and 16.5 on the
ACT.
6. (a) A plasma TV manufacturer determines that their TV’s are problem free for an average of
60 months. Assume the problem free period is normally distributed with a standard deviation
of 4 months. How many months should an extended warranty last if the company wants a TV
to have a 95% chance of being problem free during the warranty period.
(b) If the company offered a 48 month warranty, what percentage of the TV’s would be problem
free during the the warranty period?
Answer. (a) For this, 95% of normal curve must lie to the right of z, or 5% lies to the left of
z. Thus z = −1.645 and so
x = 60 − 1.645(4) = 53.42
Thus the extended warranty could last 53 months.
(b)
48 − 60
P (x > 48) = P z >
4
= P (z > −3) = 1 − .0013 = .9987
That is, 99.87% of the manufacturer’s plasma TVs will be problem free during the 48 month
warranty period.