The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. Find the probability that she has no children.

e. Find the probability that she has fewer children than the Japanese average.

f. Find the probability that she has more children than the Japanese average.

In a large population, the Poisson distribution is used to characterize the distribution of unusual events.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of one children, Japanese woman has.

Make the list of values that we want to use X may take on.

As we can see there is an upper bound for the situation at hand so,

$X=0,1,2,3,4$

The random variable X refers to the number of trials before the first success. Each trial is independent of others and has similar probability of success.

This implies that random variable X follows Poisson Distribution.

Thus, the distribution of X is$X~P\hspace{0.17em}(1.37)$

The probability that a japanese woman has no children is:

$$\begin{gathered} P(X=0)=\frac{1.{37}^0}{0!}{e}^{-1.37} \\ P(X=0)=0.25411 \end{gathered}$$

We are given that The average number of children a Japanese woman has in her lifetime is 1.37. The probability that she has fewer children than the Japanese average is

$$\begin{gathered} P(X<1.37)=P(X\le 1) \\ P(X<1.37)=P(X=0)+P(X=1) \\ P(X<1.37)=0.25411+0.34813 \\ P(X<1.37)=0.60224 \end{gathered}$$

The probability that she has more children than the Japanese average is:

$$\begin{gathered} P(X>1.37)=1-P(X\le 1.37) \\ P(X>1.37)=1-0.60224 \\ P(X>1.37)=0.39776 \end{gathered}$$