Question

The watermelons grown on a farm have normally distributed weights, with a mean weight of 80.1 ounces and a standard deviation of 17.7 ounces. Which weight interval centered at the mean includes 50% of all the watermelons?

A) [65.2, 95.0]
B) [68.2, 92.0]
C) [71.0, 89.2]
D) [74.0, 86.2]

Answer 1

The correct answer is B) [68.2, 92.0]

Explanation:

This is from usatestprep:

Given a normally distributed random variable X with mean and standard deviation , invNorm (p, , ) = x: P(X < x) = p.Because of the symmetry of the normal distribution about the mean, the bounds of the interval (x1, x2) centered at the mean, such that P(x1 < X < x2) = a arex2 = invNorm (0.5 +a2, , ); x1 = - (x2 - )Here x2 = invNorm (0.5 +0.502, 80.1, 17.7) = 92.04x1 = 80.1 - (92.04 - 80.1) = 68.06Rounded to the nearest tenth, the required interval is [68.1, 92.0].