| Number of people | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| Household probability | $0.25$ | $0.32$ | $0.17$ | $0.15$ | $0.07$ | $0.03$ | $0.01$ |
| Family probability | $0$ | $0.42$ | $0.23$ | $0.21$ | $0.09$ | $0.03$ | $0.02$ |
Let H= the number of people in a randomly selected U.S. household and F= the number of people in a randomly chosen U.S. family. Find the expected value of each random variable. Explain why this difference makes sense.
This question aims to find the expected values of the given random variables.
A random variable can be regarded as a conceptualization of a quantity whose value is determined by a random event. It is also known as the random quantity or a stochastic variable. It is a mapping or function from possible events in a sample space to a measurable space, which is frequently real numbers.
In probability and statistical analysis, the expected value is computed by adding the product of each possible outcome with its probability of occurrence. By determining expected values, investors can select the type of situation that is highly probable to accomplish a specific objective. It is a concept based on finance. In finance, it denotes the expected future value of an investment. The expected value of the occurrences can be calculated by calculating the likelihood of possible outcomes. The term is commonly used in conjunction with multivariate models and scenario analysis. It is closely linked to the idea of expected return.
Expert Answer
Let $x$ be the number of people, $p_h$ be the probability of household and $p_f$ be the probability of family, then:
| $x$ | $p_h$ | $p_f$ | $xp_h$ | $xp_f$ |
| $1$ | $0.25$ | $0$ | $0.25$ | $0$ |
| $2$ | $0.32$ | $0.42$ | $0.64$ | $0.84$ |
| $3$ | $0.17$ | $0.23$ | $0.51$ | $0.69$ |
| $4$ | $0.15$ | $0.21$ | $0.60$ | $0.84$ |
| $5$ | $0.07$ | $0.09$ | $0.35$ | $0.45$ |
| $6$ | $0.03$ | $0.03$ | $0.18$ | $0.18$ |
| $7$ | $0.01$ | $0.02$ | $0.07$ | $0.14$ |
| $\sum x p_h=2.6$ | $\sum x p_f=3.14$ |
Let $E_1$ be the expected value of the household then:
$E_1=\sum x p_h=2.6$
Let $E_2$ be the expected value of the family then:
$E_2=\sum x p_f=3.14$
The average number of people in a family is higher than the average number of people in a household, which makes sense given that all families have at least two people and all households have at least one person.
Example
A factory manufactures chairs. $2$ out of every $40$ chairs is defective, but the factory only knows when a customer complains. Assume that the factory gains a profit of $\$ 4$ on every chair sold, but loses $\$ 75$ on every defected chair since it needs to be repaired. Determine the expected profit of the factory.
Solution
Total chairs are $40$.
Defected chairs are $2$.
So the number of non-defective chairs is: $40-2=38$
Probability of non-defective chairs: $\dfrac{38}{40}$
Probability of defective chairs: $\dfrac{2}{40}$
Let $E(X)$ be the expected profit then:
$E(X)=4\left(\dfrac{38}{40}\right)+(-75)\left(\dfrac{2}{40}\right)$
$=\dfrac{19}{5}-\dfrac{15}{4}$
$=\dfrac{1}{20}$
$E(X)=0.05$
The positive expected value indicates that the factory can expect to make a profit, and the average profit per chair is $\$0.05$.