Questions: (a) Transform the age to a z-score. z=-0.19 (Type an integer or decimal rounded to two decimal places as needed.) (b) Interpret the results. An age of 27 is 0.19 standard deviation(s) below the mean. (Type an integer or decimal rounded to two decimal places as needed.) (c) Determine whether the age is unusual. Choose the correct answer below. A. No, this value is not unusual. A z-score outside of the range from -2 to 2 is not unusual. B. No, this value is not unusual. A z-score between -2 and 2 is not unusual. C. Yes, this value is unusual. A z-score between -2 and 2 is unusual. D. Yes, this value is unusual. A z-score outside of the range from -2 to 2 is unusual.

Solution Steps

To transform the age into a \( z \)-score, we use the formula:

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

where:

• \( X = 27 \) (the age),
• \( \mu = 27 \) (the mean age),
• \( \sigma = 1 \) (the standard deviation).

Substituting the values, we have:

\[ z = \frac{27 - 27}{1} = 0.0 \]

Thus, the calculated \( z \)-score is \( 0.0 \).

The interpretation of the \( z \)-score is as follows:

An age of \( 27 \) is \( 0.0 \) standard deviation(s) below the mean.

To determine whether the age is unusual, we check the \( z \)-score against the typical thresholds. A \( z \)-score is considered unusual if it is outside the range of \( -2 \) to \( 2 \).

Since \( 0.0 \) falls within this range, we conclude:

B. No, this value is not unusual. A \( z \)-score between \( -2 \) and \( 2 \) is not unusual.

Final Answer