Questions: Suppose that the mean cranial capacity for men is 1120 cc (cubic centimeters) and that the standard deviation is 300 cc. Assuming that men's cranial capacities are normally distributed, complete the following statements. (a) Approximately [] % of men have cranial capacities between 820 cc and 2020 cc. (b) Approximately 95% of men have cranial capacities between [] cc and [] cc.

Solution Steps

To find the percentage of men with cranial capacities between \( 820 \, \text{cc} \) and \( 2020 \, \text{cc} \), we first calculate the cumulative distribution function (CDF) values at these points:

• \( P(X \leq 820) \approx 0.1587 \)
• \( P(X \leq 2020) \approx 0.9987 \)

The percentage of men with cranial capacities in this range is given by:

\[ P(820 < X < 2020) = P(X \leq 2020) - P(X \leq 820) \approx 0.9987 - 0.1587 = 0.84 \]

Thus, approximately \( 84.00\% \) of men have cranial capacities between \( 820 \, \text{cc} \) and \( 2020 \, \text{cc} \).

To determine the cranial capacities that encompass approximately \( 95\% \) of men, we first find the Z critical value for a \( 95\% \) confidence level:

\[ Z = 1.96 \]

Using this Z value, we can calculate the lower and upper bounds of cranial capacities:

\[ \text{Lower Bound} = \mu - Z \cdot \sigma = 1120 - 1.96 \cdot 300 \approx 532.00 \, \text{cc} \] \[ \text{Upper Bound} = \mu + Z \cdot \sigma = 1120 + 1.96 \cdot 300 \approx 1708.00 \, \text{cc} \]

Thus, approximately \( 95\% \) of men have cranial capacities between \( 532.00 \, \text{cc} \) and \( 1708.00 \, \text{cc} \).

Final Answer