Questions: Question 10 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 2.7 hours per day unsupervised. c. What percent of the children spend over 2.3 hours per day unsupervised. % (Round to 2 decimal places) d. 89% of all children spend at least how many hours per day unsupervised? hours.

Solution Steps

The random variable \(X\), representing the amount of time a 4-year-old child on Mercury spends unsupervised per day, is normally distributed. The mean \(\mu\) is 2.9 hours, and the standard deviation \(\sigma\) is 1.4 hours. Therefore, the distribution of \(X\) is:

\[ X \sim N(2.9, 1.4^2) \]

To find the probability that a child spends less than 2.7 hours unsupervised, we calculate:

\[ P(X < 2.7) = \Phi\left(\frac{2.7 - 2.9}{1.4}\right) \]

The Z-score is:

\[ Z = \frac{2.7 - 2.9}{1.4} = -0.1429 \]

Using the standard normal distribution table or a calculator, we find:

\[ P(X < 2.7) = \Phi(-0.1429) = 0.4432 \]

To find the percentage of children spending more than 2.3 hours unsupervised, we calculate:

\[ P(X > 2.3) = 1 - P(X \leq 2.3) = 1 - \Phi\left(\frac{2.3 - 2.9}{1.4}\right) \]

The Z-score is:

\[ Z = \frac{2.3 - 2.9}{1.4} = -0.4286 \]

Using the standard normal distribution table or a calculator, we find:

\[ P(X > 2.3) = 1 - \Phi(-0.4286) = 0.3341 \]

Thus, the percentage of children spending over 2.3 hours is:

\[ 33.41\% \]

Final Answer