Question

According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 65 inches and a standard deviation of 2.5 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that the person is between 61.6 and 63.4 inches.

c. The middle 30% of Martian heights lie between what two numbers?
Low:  inches
High:  inches

Answer 1

Given that, mean (μ) = 65 inches and

standard deviation = 2.5 inches

a) X ~ N(65, 2.5)

b) We want to find, P(61.6 < X < 63.4)

Therefore, required probability is 0.1742

c) We want to find, values of x1 and x2 such that, P(x1 < X < x2) = 0.30

First we find, z-score such that, P(-z < Z < z) = 0.30

P(-z < Z < z) = 0.30

=> 2 * P(Z < z) - 1 = 0.30

=> 2 * P(Z < z) = 1.30

=> P(Z < z) = 0.65

Using Excel we get, z-score corresponding area under the normal curve of 0.65 is, z = 0.38532

Excel Command : NORMSINV (0.65) = 0.38532

For z = -0.38532

x1 = (-0.38532 * 2.5) + 65 = -0.9633 + 65 = 64.0367

For x2 = (0.38532 * 2.5) + 65 = 0.9633 + 65 = 65.9633

=> P(64.0367 < X < 65.9633) = 0.30

Hence, we get

Low = 64.0367 inches

High = 65.9633 inches

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