Questions: Multiple Choice. The heights of adults in one town have a mean of 66.8 inches and a standard deviation of 3.5 inches. What is the percentage of adults in this town have heights between 56.3 and 77.3 inches? Circle your answer. (a) The percentage is at most 75% (b) The percentage is at least 75% (c) The percentage is at most 89% (d) The percentage is at least 89%

Solution Steps

To find the percentage of adults with heights between 56.3 and 77.3 inches, we can use the properties of the normal distribution. We will calculate the z-scores for the given heights and then use the cumulative distribution function (CDF) to find the probabilities. The difference between these probabilities will give us the percentage of adults within this range.

To find the percentage of adults with heights between 56.3 and 77.3 inches, we first calculate the z-scores for these heights using the formula:

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

For the lower bound (56.3 inches):

\[ z_{\text{lower}} = \frac{56.3 - 66.8}{3.5} = -3.0 \]

For the upper bound (77.3 inches):

\[ z_{\text{upper}} = \frac{77.3 - 66.8}{3.5} = 3.0 \]

Using the cumulative distribution function (CDF) for a standard normal distribution, we find the probabilities corresponding to these z-scores.

For \( z_{\text{lower}} = -3.0 \):

\[ P(Z < -3.0) = 0.001349898031630093 \]

For \( z_{\text{upper}} = 3.0 \):

\[ P(Z < 3.0) = 0.9986501019683699 \]

The percentage of adults with heights between 56.3 and 77.3 inches is the difference between these probabilities:

\[ P(56.3 < X < 77.3) = P(Z < 3.0) - P(Z < -3.0) = 0.9986501019683699 - 0.001349898031630093 \]

\[ = 0.9973002039367398 \]

Converting this to a percentage:

\[ 99.73\% \]

Final Answer