(part 1)

a. Let X = the amount of time, in hours, a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.

Find P(1.8 < x < 2.75).

First, calculate the z-scores for each x-value.

$\begin{array}{l}z=\frac{1.8-2}{0.5}=\frac{-0.2}{0.5}=-0.40\hfill \\ z=\frac{2.75-2}{0.5}=\frac{0.75}{0.5}=1.5\hfill \end{array}$

Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores.

The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or $0.9332-0.3446$, which equals 0.5886.

normalcdf(1.8,2.75,2,0.5) = 0.5886

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.