Question

suppose \average\ is defined operationally as the middle 4 percent of a distribution. assume that womens heights are normally distributed with a mean of 162 cm and a standard deviation of 6 cm. what is the tallest that a women can be and still be considered to be of \average\ height?

Explanation:

Step1: Find tail percentages

The middle 4% leaves $100\% - 4\% = 96\%$ in the two tails, so each tail is $\frac{96\%}{2} = 48\%$. The upper bound of the middle 4% corresponds to the $50\% + 2\% = 52\%$ percentile (since 2% is half of the middle 4%).

Step2: Find z-score for 52nd percentile

Using a z-table or calculator, the z-score corresponding to a cumulative probability of 0.52 is approximately $z = 0.05$.

Step3: Calculate height with z-score formula

The z-score formula is $z = \frac{x - \mu}{\sigma}$, rearranged to $x = \mu + z\sigma$. Substitute $\mu = 162$, $z = 0.05$, $\sigma = 6$:
$x = 162 + (0.05)(6)$

Step4: Compute final value

$x = 162 + 0.3 = 162.3$

Answer:

162.3 cm