suppose \average\ is defined operationally as the middle 4 percent of a distribution. assume that womens…

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?

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?

Answer

Explanation:

Step1: Find upper tail percentage

The middle 4% leaves $\frac{100% - 4%}{2} = 98%$ of the distribution below the upper bound of "average".

Step2: Find z-score for 0.98 cumulative probability

Using standard normal distribution tables or a calculator, the z-score corresponding to a cumulative probability of 0.98 is approximately $z = 2.05$.

Step3: Calculate upper height bound

Use the z-score formula $z = \frac{x - \mu}{\sigma}$, rearranged to $x = \mu + z\sigma$. Substitute $\mu = 162$, $\sigma = 6$, $z = 2.05$: $x = 162 + 2.05 \times 6$ $x = 162 + 12.3 = 174.3$

Answer:

174.3 cm