Question

Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.

60% of the children spend at least how long per day unsupervised? (Round your answer to two decimal places.)

Answer 1

Solution :  

= 3

= / n = 1.8 / 4 = 0.9

P(Z > -0.25) = 0.60

z = -0.25

= z * + = -0.25 * 0.9 + 3 = 2.78

Answer = 2.78 hours