Question

4. In China, six-year old children average three hours per day of unsupervised activity (not including sleep/rest times). Most of the unsupervised children live in very rural areas that are considered to be quite safe. Suppose that the time spent alone is normally distributed with a standard deviation of 0.5 hours. Based on this information, … b. What is the probability that a child in this study will spend between 2.5 and 3.5 hours per day in unsupervised activity? c. What is the probability that a child will spend less than 2 hours and 15 minutes per day in unsupervised activity? d. What is the probability that a child will spend more than 4 hours 10 minutes per day in unsupervised activity? e. How long would a child have to remain unsupervised in this study for him/her to be 3.05 standard deviations above the mean? How uncommon would this be?

Answer 1

Here

b) We have to find

                                       

                                        = P(-1 < Z < 1)

                                        = P(Z < 1) - P(Z < -1)

                                        = 0.8413 - 0.1587                   (From statistical table of Z values)

                                        = 0.6826

Probability that child will spend between 2.5 and 3.5 hours per day in unsupervised activity is 0.6826

c)

                             

                              = P(Z < -1.7)

                              = 0.1423                                     (From statistical table of Z values)

Probability that child will spend less than 2 hours and 15 minutes per day in unsupervised is activity is 0.1423.

d)

                            

                             = P(Z > 2.2)

                            = 1 - P(Z < 2.2)

                            = 1 - 0.9861            (From statistical table of Z values)

                            = 0.0139

Probability that child will spend more than 4 hours and 10 minutes per day in unsupervised is activity is 0.0139

e)

      = 3 + 3.05 * 0.5

      = 3 + 1.525

     = 4.525

A child would have to remain unsupervised for 4.525 hours in this study for him/her to be 3.05 standard deviations above the mean.