On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.4 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.

Required:
a. What is the distribution of X?
b. Find the probability that the child spends less than 1.4 hours per day unsupervised.
c. What percent of the children spend over 3.5 hours per day unsupervised.

Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed

Generally the probability that the child spend less than one hour in a day is mathematically represented as

[tex]P(X < 1.4) = P(\frac{X - \mu}{\sigma} < \frac{1.4 - \mu}{\sigma} )[/tex]

Here [tex]\frac{X - \mu}{\sigma } = Z (The\ standardized\ value\ of\ X)[/tex]

So

[tex]P(X < 1.4) = P(Z < \frac{1.4 - 3.0}{1.4} )[/tex]

[tex]P(X < 1.4) = P(Z < -1.1429 )[/tex]

From the z-table the value of

[tex]P(Z < -1.1429 )=0.12654[/tex]

So [tex]P(X < 1.4) = 0.12654[/tex]

Generally the percentage of children that spends over 3.5 hours unsupervised is mathematically represented as

[tex]P(X > 3.5) = P(\frac{X - \mu}{\sigma} > \frac{3.5 - \mu}{\sigma} )[/tex]

[tex]P(X > 3.5) = P(Z > \frac{3.5 - 3.0}{1.4} )[/tex]

[tex]P(X > 3.5) = P(Z > 0.3571 )[/tex]

From the z-table the value of

[tex]P(Z >0.3571 )=0.36051[/tex]

So [tex]P(X > 3.5) = 0.36051[/tex]