Question

question 9 let x be the number of students who attend an office hours session a day prior to an exam. x is known to be distributed as follows: p(x)=kx 0≤x≤5 what is the expected number of students to attend such a session? 0 6 0 15 0 55 0 4

Explanation:

Step1: Find the value of k

Since $\sum_{x = 0}^{5}p(x)=1$, and $p(x)=kx$ for $0\leq x\leq5$. We have $\sum_{x = 0}^{5}kx=k\sum_{x = 0}^{5}x$. Using the sum - of - arithmetic - series formula $\sum_{i = 1}^{n}i=\frac{n(n + 1)}{2}$, here $\sum_{x=0}^{5}x=\frac{5\times(5 + 1)}{2}=15$. So $k\times15 = 1$, then $k=\frac{1}{15}$.

Step2: Calculate the expected value E(X)

The formula for the expected value of a discrete random variable is $E(X)=\sum_{x}x\cdot p(x)$. Here, $p(x)=\frac{1}{15}x$ for $0\leq x\leq5$. So $E(X)=\sum_{x = 0}^{5}x\cdot\frac{1}{15}x=\frac{1}{15}\sum_{x = 0}^{5}x^{2}$. Using the formula $\sum_{i = 1}^{n}i^{2}=\frac{n(n + 1)(2n + 1)}{6}$, here $\sum_{x = 0}^{5}x^{2}=\frac{5\times(5 + 1)\times(2\times5+1)}{6}=55$. Then $E(X)=\frac{1}{15}\times55=\frac{11}{3}\approx3.67$. But if we assume there is a calculation error above and recalculate:
The expected value formula $E(X)=\sum_{x}x\cdot p(x)$. Since $p(x)=kx$ and $k=\frac{1}{15}$, $E(X)=\sum_{x = 0}^{5}x\cdot\frac{1}{15}x=\frac{1}{15}\sum_{x=0}^{5}x^{2}=\frac{1}{15}\times55=\frac{11}{3}\approx 3.67$. If we consider the correct way:
$E(X)=\sum_{x = 0}^{5}x\cdot p(x)=\sum_{x = 0}^{5}x\cdot kx=k\sum_{x = 0}^{5}x^{2}$. After finding $k = \frac{1}{15}$ and $\sum_{x=0}^{5}x^{2}=55$, $E(X)=\frac{1}{15}\times55=\frac{11}{3}\approx3.67$. The closest integer value among the options is 4.

Answer:

4