Question

In China, four-year-olds average three 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 select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.

a. In words, define the random variable X.

b. X ~ _____(_____,_____)

c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.

d. What percent of the children spend over ten hours per day unsupervised?

e. Seventy percent of the children spend at least how long per day unsupervised?

Short answer

1. $X=$The amount of time (in hours) that Chinese four-year-olds who live in rural areas spend alone.
2. $X~N(3,1.5)$
   
3. The chances that the youngster spends less than one hour per day unsupervised is $\mathrm{P}(\mathrm{x}<1)=0.0912\text{.}$
   
4. The percentage of youngsters who spend more than ten hours per day alone is 0.00015.
5. The ${70}^{th}$percentile of the children spend at least$2.21$hours per day unsupervised.

Step-by-step solution

Part(a) Step 1: Given Information 

Given in the question that, in China four-year-olds average three 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.

Part(a) Step 2: Explanation 

A random variables are the variables that represents the probability in a random experiment. Upper case alphabets, such as $X,Y$, and so on, are used to represent it. The problem at hand entails calculating the quantity of time (in hours) spent alone by Chinese four-year-olds in rural locations. In a particular interval, the quantity of time varies for each child. As a result, it can be thought of as a random variable with the following definition:

$X$: The amount of time (in hours) that Chinese four-year-olds who live in rural areas spend alone.$X$ is a continuous random variable because it can take an endless number of values in a given interval.

Part(b) Step 1: Given Information 

We have to fill$X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_)$

Part(b) Step 2: Explanation 

$X$is a random variable with a mean of 3 and a standard deviation of $1.5$that represents the amount of time (in hours) spent alone by Chinese four-year-olds living in rural areas.

If a random variable $X$has a normal distribution with a mean of $\mu$and a standard deviation of $s$, it is denoted as $X~N(\mu ,s)$

As a result, the random variable $X$in this situation is denoted as:$X~N(3,1.5)$

Part(c) Step 1: Given Information 

Given that, $X~N(3,1.5)$we must calculate the chance that the child spends less than one hour every day unaccompanied.

Part(c) Step 2: Explanation 

The probability that the youngster spends less than one hour unsupervised per day is calculated as follows:

$$\begin{gathered} P(x<1)=P\left(\frac{x-\mu }{\sigma }<\frac{1-3}{1.5}\right) \\ =P(z<-1.3333) \end{gathered}$$

The shaded zone in the graph below depicts the needed probability. As a result, it can be written:

$P(z<-1.3333)=0.5-P(-1.3333\le z\le 0)$

$=0.5$

Due to symmetry:

$-P(0\le z\le 1.3333)=0.5$

From normal table $-0.4088=0.0912$

Part(d) Step 1: Given Information 

When the amount of time $\left(X\right)$ spent alone by Chinese four-year-olds living in a rural location is such that $X~N(3,1.5)$, the percentage of children who spend more than 10 hours each day unattended.

Part(d) Step 2: Explanation 

The probability that the youngsters spend more than ten hours each day unsupervised is calculated as follows:

$P(x>10)=P\left(\frac{x-\mu }{\sigma }>\frac{10-3}{1.5}\right)=P(z>4.6667)$

The shaded region in the graph below depicts the needed probability, which is calculated as

$$\begin{gathered} P(z>4.6667)=0.5-P(0\le z\le 4.6667) \\ =0.5-0.499998 \\ =0.0000015 \end{gathered}$$

Therefore, the required percentage is :$50.0000015\times 100=\mathbf{0}\mathbf{.}\mathbf{00015}$

Part (e) Step 1: Given Information 

Given in the question that,$X~N(3,1.5)$

Part (e) Step 2: Explanation 

The needed probability is calculated as follows:

$P(x<k)=P\left(\frac{x-\mu }{\sigma }<\frac{k-3}{1.5}\right)=P\left(z<z_1\right)=0.3$

We get the following results from the normal distribution tables:

$z_1=-0.5244$

In the equation $x=\mu +z\sigma$replace $x$with $k$and $z$with $z_1$.

$k=3+(-0.5244)*1.5=2.21$