Question

A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distributed, what is the z score for students being active 48 minutes per week?a. 1.0b. -1.0c. 0d. There is not enough information to answer this question.

Answer 1

Answer:

The correct option is (b).

Step-by-step explanation:

If X $\sim$ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z $\sim$ N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

μ = 60 minutes

σ = 12 minutes

Compute the z-score for the student being active 48 minutes as follows:

$Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0$

Thus, the z-score for the student being active 48 minutes is -1.0.

The correct option is (b).