On the planet of Mercury, 4-year-olds average 3.2 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 $X ? X \sim N(\square, \square$, b. Find the probability that the child spends less than 3.4 hours per day unsupervised.

Step 1 ：The problem is asking for two things. First, it wants to know the distribution of the variable $X$, which represents the amount of time a child spends alone per day. This is a normal distribution, as stated in the problem, with a mean of 3.2 hours and a standard deviation of 1.5 hours. So, the distribution of $X$ is $N(3.2, 1.5^2)$.

Step 2 ：Second, it wants to find the probability that a child spends less than 3.4 hours per day unsupervised. To find this, we need to calculate the z-score for 3.4 hours, and then find the area under the normal distribution curve to the left of this z-score.

Step 3 ：The z-score is calculated as $(x - \mu) / \sigma$, where $x$ is the value we're interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = (3.4 - 3.2) / 1.5 = 0.1333$.

Step 4 ：Once we have the z-score, we can use a z-table or a function like scipy's norm.cdf() to find the cumulative distribution function (CDF) up to that z-score, which gives us the probability we're looking for. The probability is approximately 0.5530.

Step 5 ：Final Answer: a. The distribution of $X$ is $X \sim N(3.2, 1.5^2)$. b. The probability that the child spends less than 3.4 hours per day unsupervised is approximately \boxed{0.5530}.