Questions: 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 3.5 in tall (between 51.5 inches and 63.5 inches)? Round your answer to three decimal places. Proportion =

Solution Steps

To find the proportion of 10-year-old boys whose heights are between 51.5 inches and 63.5 inches, we can use the properties of the normal distribution. We will calculate the z-scores for both height limits and then use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to these z-scores. The proportion of boys within this height range is the difference between these two probabilities.

To find the proportion of boys with heights between 51.5 inches and 63.5 inches, we first calculate the z-scores for these height limits using the formula:

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

For the lower bound (51.5 inches):

\[ z_{\text{lower}} = \frac{51.5 - 55.5}{2.7} \approx -1.4815 \]

For the upper bound (63.5 inches):

\[ z_{\text{upper}} = \frac{63.5 - 55.5}{2.7} \approx 2.963 \]

Next, we find the cumulative probabilities for these z-scores using the standard normal distribution:

\[ P(Z < z_{\text{lower}}) \approx 0.06924 \]

\[ P(Z < z_{\text{upper}}) \approx 0.9985 \]

The proportion of boys with heights between 51.5 inches and 63.5 inches is the difference between these cumulative probabilities:

\[ \text{Proportion} = P(Z < z_{\text{upper}}) - P(Z < z_{\text{lower}}) \approx 0.9985 - 0.06924 = 0.9292 \]

Final Answer