The following table shows the approximate numbers of school goers in the U.S. (residents who attended some educational institution) in 1998, broken down by age group. Use the rounded midpoints of the given measurement classes to compute the probability distribution of the age X of a school goer. (Round probabilities to two decimal places.) Hence compute the expected value of X. $$\begin{array} { c c c c c c c }
\text { Age } & 3 - 6.9 & 7 - 12.9 & 13 - 16.9 & 17 - 22.9 & 23 - 26.9 & 27 - 42.9 \
\begin{array} { c }
\text { Population } \
\text { (Millions) }
\end{array} & 11 & 22 & 16 & 15 & 4 & 7
\end{array}$$
A) $E ( X ) = 15.2$
B) $E ( X ) = 0.99$
C) $E ( X ) = 7.65$
D) $E ( X ) = 12.05$
E) $E ( X ) = 10.8$

Question

The following table shows the approximate numbers of school goers in the U.S. (residents who attended some educational institution) in 1998, broken down by age group. Use the rounded midpoints of the given measurement classes to compute the probability distribution of the age  X of a school goer. (Round probabilities to two decimal places.) Hence compute the expected value of X. $$\begin{array} { c c c c c c c } 
\text { Age } & 3 - 6.9 & 7 - 12.9 & 13 - 16.9 & 17 - 22.9 & 23 - 26.9 & 27 - 42.9 \
\begin{array} { c } 
\text { Population } \
\text { (Millions) }
\end{array} & 11 & 22 & 16 & 15 & 4 & 7
\end{array}$$  
A) $E ( X ) = 15.2$
B) $E ( X ) = 0.99$
C) $E ( X ) = 7.65$
D) $E ( X ) = 12.05$
E) $E ( X ) = 10.8$

Quizplus · Accepted Answer

To calculate the expected value $E(X)$ of the age $X$ of a school goer, we first need to find the probability distribution of $X$, and then use these probabilities to compute the expected value. The midpoint for each age group is calculated as the average of the lower and upper bounds of the age group. Then, we find the probability of each age group by dividing the population of the group by the total population. Finally, we multiply each midpoint by its corresponding probability and sum these products to get the expected value.1. Calculate midpoints for each age group:- $3 - 6.9$: midpoint = $\frac{3 + 6.9}{2} = 4.95$- $7 - 12.9$: midpoint = $\frac{7 + 12.9}{2} = 9.95$- $13 - 16.9$: midpoint = $\frac{13 + 16.9}{2} = 14.95$- $17 - 22.9$: midpoint = $\frac{17 + 22.9}{2} = 19.95$- $23 - 26.9$: midpoint = $\frac{23 + 26.9}{2} = 24.95$- $27 - 42.9$: midpoint = $\frac{27 + 42.9}{2} = 34.95$2. Calculate the total population:Total population = $11 + 22 + 16 + 15 + 4 + 7 = 75$ million3. Calculate the probability of each age group:- $3 - 6.9$: $P = \frac{11}{75}$- $7 - 12.9$: $P = \frac{22}{75}$- $13 - 16.9$: $P = \frac{16}{75}$- $17 - 22.9$: $P = \frac{15}{75}$- $23 - 26.9$: $P = \frac{4}{75}$- $27 - 42.9$: $P = \frac{7}{75}$4. Calculate the expected value $E(X)$:$$E(X) = 4.95\left(\frac{11}{75}\right) + 9.95\left(\frac{22}{75}\right) + 14.95\left(\frac{16}{75}\right) + 19.95\left(\frac{15}{75}\right) + 24.95\left(\frac{4}{75}\right) + 34.95\left(\frac{7}{75}\right)$$After performing the calculations and rounding to two decimal places, the expected value $E(X)$ is approximately 15.2.

The Following Table Shows the Approximate Numbers of School Goers