Questions: Question 17 of 17 - / 2 Boys Heights Heights of ten year old boys (5th graders) follow an approximate normal distribution with mean μ=55.5 inches and standard deviation σ=2.7 inches. (a) According to this normal distribution, what proportion of 10-year-old boys are between 4 ft 3.5 in and 5 ft 6.5 in tall (between 51.5 inches and 66.5 inches)? Round your answer to three decimal places. Proportion = (b) A parent says his 10-year-old son is in the 99th percentile in height. How tall is this boy? Round your answer to two decimal places. Height = i inches

Solution Steps

To find the proportion of 10-year-old boys whose heights are between 51.5 inches and 66.5 inches, we use the cumulative distribution function (CDF) of the normal distribution:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Where:

• \( Z_{end} \) corresponds to 66.5 inches
• \( Z_{start} \) corresponds to 51.5 inches

After calculating the Z-scores, we find:

\[ P = \Phi(4.074) - \Phi(-1.481) = 0.931 \]

Thus, the proportion of boys between 51.5 inches and 66.5 inches is:

\[ \text{Proportion} = 0.931 \]

To determine the height of a boy in the 99th percentile, we first calculate the Z-score corresponding to this percentile:

\[ z = \frac{X - \mu}{\sigma} \]

Where:

• \( X \) is the height we want to find
• \( \mu = 55.5 \) inches (mean)
• \( \sigma = 2.7 \) inches (standard deviation)

Using the Z-score for the 99th percentile, we find:

\[ z = \frac{0.99 - 55.5}{2.7} = -20.19 \]

Now, we can find the height corresponding to the 99th percentile using the Z-score:

\[ X = \mu + z \cdot \sigma \]

Substituting the values:

\[ X = 55.5 + (-20.19) \cdot 2.7 = 0.99 \text{ inches} \]

Final Answer