Answer:
The correct option is (b).
Step-by-step explanation:
If X [tex]\sim[/tex] N (µ, σ²), then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z [tex]\sim[/tex] 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:
[tex]Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0[/tex]
Thus, the z-score for the student being active 48 minutes is -1.0.
The correct option is (b).
