Questions: In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 53.1 inches, and standard deviation of 5.6 inches. A) What is the probability that a randomly chosen child has a height of less than 59.9 inches? Answer = (Round your answer to 3 decimal places.) B) What is the probability that a randomly chosen child has a height of more than 42.5 inches? Answer = (Round your answer to 3 decimal places.)

Solution Steps

We are given that the height measurements of ten-year-old children in the United States of Heightlandia are approximately normally distributed with a mean (\(\mu\)) of 53.1 inches and a standard deviation (\(\sigma\)) of 5.6 inches. We need to find the probabilities for two scenarios:

A) The probability that a randomly chosen child has a height of less than 59.9 inches.

B) The probability that a randomly chosen child has a height of more than 42.5 inches.

For part A, we need to find \(P(X < 59.9)\), where \(X\) is the height of a randomly chosen child. This can be calculated using the cumulative distribution function (CDF) of the normal distribution.

1. Calculate the Z-score for 59.9 inches: \[ Z = \frac{59.9 - 53.1}{5.6} = 1.214 \]

2. Use the standard normal distribution to find \(P(Z < 1.214)\). This is equivalent to \(\Phi(1.214)\).

3. Since the lower bound is negative infinity, the probability is: \[ P(X < 59.9) = \Phi(1.214) - \Phi(-\infty) = 0.888 \]

For part B, we need to find \(P(X > 42.5)\).

1. Calculate the Z-score for 42.5 inches: \[ Z = \frac{42.5 - 53.1}{5.6} = -1.893 \]

2. Use the standard normal distribution to find \(P(Z > -1.893)\). This is equivalent to \(1 - \Phi(-1.893)\).

3. Since the upper bound is positive infinity, the probability is: \[ P(X > 42.5) = \Phi(\infty) - \Phi(-1.893) = 0.971 \]

Final Answer