Question

On the planet of Mercury, 4-year-olds average 2.9 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N( , )

b. Find the probability that the child spends less than 2.7 hours per day unsupervised.  

c. What percent of the children spend over 4.1 hours per day unsupervised ? % (Round to 2 decimal places)

d. 71% of all children spend at least how many hours per day unsupervised?  hours.

Answer 1

a) The distribution of X is normal.

b)Since the distribution is normal then using Z score probability is calculated

Hence P(X<2.7) =P(Z<-1.33)

P-value is calculated using the Z table shown below as

=1-0.4471

=0.5529.

c) At 4.1 Z score is

Hence P(X>4.1)=P(Z>0.8)

=0.2119.

d) At highest 71% of the unsupervised hours, the Z score is 0.55 hence using the Z score

hence 71 % 0f them spend at least 1.7 days unsupervised.

The Z scores and P- values are calculated using Z table shown below as