find the probability that the hours spent on the home computer by the participant are more than 9.4 hours per week

Question

the mean is 7 hours and the standard deviation is 1 hour

Answer 1

Assuming a normal distribution we can calculate a Z score=(X-m)/s, where X=9.4hr, m=pop mean=7hr, s=pop standard deviation=1hr.

Z=(9.4-7)/1=2.4.

p(≤2.4)=99.18% (from tables).

So p(≥2.4)=100-99.18=0.82%.