Chapter 6, Problem 52CP

Grandparents Let X = the number of living grandparents that a randomly selected adult American has. According to recent General Social Surveys, its probability distribution is approximately P(0) = 0.71, P(1) = 0.15, P(2) = 0.09, P(3) = 0.03, P(4) = 0.02.

1. a. Does this refer to a discrete or a continuous random variable? Why?
2. b. Show that the probabilities satisfy the two conditions for a probability distribution.
3. c. Find the mean of this probability distribution.

To determine

Identify whether the random variable X is discrete or continuous.

Answer to Problem 52CP

The random variable X is discrete.

Explanation of Solution

The given information is that the random variable X is the number of living grandparents of a randomly selected American adult. From the survey, the probability distribution is approximately $P\left(0\right)=0.71,P\left(1\right)=0.15,P\left(2\right)=0.09,P\left(3\right)=0.03$ and $P\left(4\right)=0.02$.

Discrete random variable:

If the random variables, consider a set of distinct possible values for the trial then it is called discrete random variable.

Continuous random variable:

If the random variables, consider the possible values and it form an interval then it is called continuous random variable

In the given situation, the random variable X represents the number of grandparents and it will be a distinct number because the number of grandparents does not be in an interval.

Thus, the random variable X is discrete.

To determine

Show that the probability values satisfy the probability distribution conditions.

Explanation of Solution

Conditions for the probability distribution:

1. 1. For every value of x, the value of probability will lie between 0 and 1.
2. 2. The total of all probabilities for the possible values of x is equal to 1.

From the given information, the all probability values lie between 0 and 1.

The sum of probabilities is,

$\begin{array}{c}P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)=0.71+0.15+0.09+0.03+0.02\\ =1\end{array}$

Therefore, the total sum of probability distribution is 1.

Thus, the probability values satisfy the probability distribution conditions.

To determine

Calculate the mean of the probability distribution.

Answer to Problem 52CP

The mean of the probability distribution is 0.50.

Explanation of Solution

Calculation:

The formula for mean is,

$\mu ={\displaystyle \sum xP\left(x\right)}$

From the probability distribution, calculate the mean value for number of grandparents is,

$\begin{array}{c}\mu =\left(0\times P\left(0\right)\right)+\left(1\times P\left(1\right)\right)+\left(2\times P\left(2\right)\right)+\left(3\times P\left(3\right)\right)+\left(4\times P\left(4\right)\right)\\ =\left(0\times 0.15\right)+\left(1\times 0.15\right)+\left(2\times 0.09\right)+\left(3\times 0.03\right)+\left(4\times 0.02\right)\\ =0+0.15+0.18+0.09+0.08\\ =0.50\end{array}$

Thus, the mean value is 0.50.