# Solved (Free): The Employee Benefit Research Institute reports that 69% of workers reported that they and/or their spouse had saved some money for retirement

#### ByDr. Raju Chaudhari

Mar 28, 2021

The Employee Benefit Research Institute reports that 69% of workers reported that they and/or their spouse had saved some money for retirement. If a random sample of 10 workers is taken, then determine the

a. Expected number of workers and/or their spouse have saved some money for retirement
b. Standard deviation of the workers and/or their spouse have saved some money for retirement
c. Probability that exactly 2 works and/or their spouse have saved some money for retirement
d. Probability that fewer than 7 workers and/or their spouse have saved some money for retirement.

#### Solution

Here $X$ denote the number of workers reported that they and/or their spouse have saved some money for retirementout of a sample of 10.

$p$ be the percent of the workers and/or their spouse had saved some money for retirement.

Given that $p=0.69$ and $n =10$. Thus$X\sim B(10, 0.69)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{10}{x} (0.69)^x (1-0.69)^{10-x},\\ &\quad x=0,1,\cdots, 10. \end{aligned}

a. Expected number of workers and/or their spouse have saved some money for retirement is

 \begin{aligned} E(X) &= np\\ &= 10\times 0.69\\ &= 6.9 \end{aligned}

b. Standard deviation of workers and/or their spouse have saved some money for retirement is

 \begin{aligned} sd(X) &= \sqrt{np(1-p)}\\ &= \sqrt{10\times 0.69\times (1- 0.69)}\\ &= \sqrt{2.139}\\ &= 1.4625 \end{aligned}

c. The probability that exactly 2 works and/or their spouse have saved some money for retirement

 \begin{aligned} P(X=2) & =\binom{10}{2} (0.69)^{2} (1-0.69)^{10-2}\\ &=0.0018 \end{aligned}

d. The probability that fewer than 7 workers and/or their spouse have saved some money for retirement is

 \begin{aligned} P(X< 7) & =1- P(X\geq 7)\\ &= 1- \sum_{x=7}^{10} P(x)\\ & =1- \bigg(P(X=7)+P(X=8)+P(X=9)+P(X=10)\bigg)\\ &= 1-\bigg(0.2662+0.2222+0.1099+0.0245\bigg) \\ &= 0.3772 \end{aligned}