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 1.4 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.a. What is the distribution of X? X ~ N(,)b. Find the probability that the child spends less than 2.6 hours per day unsupervised. c. What percent of the children spend over 2.5 hours per day unsupervised. % (Round to 2 decimal places)d. 72% of all children spend at least how many hours per day unsupervised? hours.

Answer 1

Part a.

From the given infomation, the mean is equal to

`\mu=2.9\text{ hours}`

and the standard deviation

`\sigma=1.4\text{ hours}`

Then, the distribution of X is:

`N(2.9,1.4)`

Part b.

In this case, we need to find the following probability:

`P(X<2.6)`

So, in order to find this value, we need to convert the 2.6 hours into a z-value score by means of the z-score formula:

`z=(X-\mu)/(\sigma)`

Then, by substituting the given values into the formula, we get

`\begin{gathered} z=(2.6-2.9)/(1.4) \\ z=-0.214285 \end{gathered}`

Then, the probability we must find in the z-table is:

`P(z<-0.214285)`

which gives

`P(z<-0.214285)=0.41516`

Therefore, by rounding to 4 decimal places, the answer for part b is: 0.4152

Part c.

In this case, we need to find the following probability

`P(X>2.5)`

Then, by converting 2.5 to a z-value, we have

`\begin{gathered} z=(2.5-2.9)/(1.4) \\ z=-0.285714 \end{gathered}`

So, we need to find on the z-table:

`P(z>-0.285714)`

which gives

`P(z\gt-0.285714)=0.61245`

Then, by multiplying this probability by 100% and rounding to the nearest hundreadth,

the answer for part c is: 61.25 %

Part d.

In this case, we have the following information:

`P(z>Z)=0.72`

and we need to find Z. From the z-table, we get

`Z=0.58284`

Then, from the z-value formula, we have

`-0.58284=(X-2.9)/(1.4)`

and we need to isolate the amount of hours given by X. Then, by multiplying both sides by 1.4, we obtain

`-0.815976=X-2.9`

Then, X is given by

`\begin{gathered} X=2.9-0.815976 \\ X=2.0840 \end{gathered}`

So, by rounding to 4 decimal places, the answer is: 2.0840 hours.