Download STA 120 * Quiz #4

STA 120 – Practice Quiz #4
1) Find the area under the Normal curve corresponding to each statement.
a) P(z < 0.53)
b) P(z < –1.02)
c) P(z < -5.38)
d) P(z > –2.43)
e) P(z > 0.89)
f) P(z > -5.07)
g) P(1.48 < z < 1.84)
h) P(–2.01 < z < –1.64)
i) P(–0.27 < z < 1.58)
j) P(-6.13 < z < 4.49)
2) The US Post Office says that the weights of first-class letters follows a Normal
distribution with mean 0.69 ounces and standard deviation 0.14 ounces.
a) First-class letters weighing more than 1 ounce require additional postage. What
proportion of first-class letters weigh more than 1 ounce?
b) What weight would correspond to the 80th percentile of first-class letter
weights?
3) Men’s hat sizes are related to their head circumference. Suppose that men’s head
circumferences are approximately Normally distributed with mean 56.2 cm and standard
deviation 1.82 cm. The table below lists some hat sizes and the corresponding head
circumference.
Hat Size Head Circumference (cm)
7¼
57.8 ≤ circumference < 58.7
7 3/8
58.7 ≤ circumference < 59.6
7½
59.6 ≤ circumference < 60.5
a) What percent of men require hat size 7 ¼ ?
b) What percent of men require hat size larger than 7 ½ ?
4) Suppose we have a binomial random variable, X, which is counting the number of
successes in 500 trials, and suppose the probability of success on each trial is 0.68.
a) What are the mean and standard deviation of this binomial random variable?
b) Use the Normal approximation, including the correction factor, to approximate
the probability of having between 345 and 360 successes (inclusive).
STA 120 – Practice Quiz #4 - Solutions
1) Find the area under the Normal curve corresponding to each statement.
a) P(z < 0.53) = 0.7019
b) P(z < –1.02) = 0.1539
c) P(z < -5.38) = approximately 0
d) P(z > –2.43) = 1 – 0.0075 = 0.9925
e) P(z > 0.89) = 1 – 0.8133 = 0.1867
f) P(z > -5.07) = approximately 1
g) P(1.48 < z < 1.84) = 0.9671 – 0.9306 = 0.0365
h) P(–2.01 < z < –1.64) = 0.0505 – 0.0222 = 0.0283
i) P(–0.27 < z < 1.58) = 0.9429 – 0.3936 = 0.5493
j) P(-6.13 < z < 4.49) = approximately 1
2) The US Post Office says that the weights of first-class letters follows a Normal
distribution with mean 0.69 ounces and standard deviation 0.14 ounces.
a) First-class letters weighing more than 1 ounce require additional postage. What
proportion of first-class letters weigh more than 1 ounce?
P(X > 1) = P(z > 2.21) = 1 – 0.9864 = 0.0136
b) What weight would correspond to the 80th percentile of first-class letter
weights?
80th percentile corresponds to a z score of 0.84.
X = 0.69 + 0.84(0.14) = 0.8076 ounces
3) Men’s hat sizes are related to their head circumference. Suppose that men’s head
circumferences are approximately Normally distributed with mean 56.2 cm and standard
deviation 1.82 cm. The table below lists some hat sizes and the corresponding head
circumference.
Hat Size Head Circumference (cm)
7¼
57.8 ≤ circumference < 58.7
7 3/8
58.7 ≤ circumference < 59.6
7½
59.6 ≤ circumference < 60.5
a) What percent of men require hat size 7 ¼ ?
P(57.8 < X < 58.7) = P(0.88 < z < 1.37) = 0.9147 – 0.8106 = 0.1041 = about 10%
b) What percent of men require hat size larger than 7 ½ ?
P(X > 60.5) = P(z > 2.36) = 1 – 0.9909 = 0.0091
4) Suppose we have a binomial random variable, X, which is counting the number of
successes in 500 trials, and suppose the probability of success on each trial is 0.68.
a) What are the mean and standard deviation of this binomial random variable?
n = 500, p = 0.68, q = 0.32
𝜇 = 𝑛𝑝 = 500(0.68) = 340
𝜎 = √𝑛𝑝𝑞 = √500(0.68)(0.32) = 10.43
b) Use the Normal approximation, including the correction factor, to approximate
the probability of having between 345 and 360 successes (inclusive).
P(344.5 < X < 360.5) = P(0.43 < z < 1.97) = 0.9756 – 0.6664 = 0.3092