# Solved (Free): Among companies doing highway or bridge construction, 80% test employees for substance abuse

#### ByDr. Raju Chaudhari

Apr 4, 2021

Among companies doing highway or bridge construction, 80% test employees for substance abuse (based on data from the Construction Financial Management Association) A study involves the random selection of 12 such companies.

(a) Find the probability that exactly 7 of the 12 companies test for substance abuse.

(b) Find the probability that at least half of the companies test for substance abuse.

(c) For such group of 12 companies, find the mean and standard deviation for the number (among 12) that test for substance abuse.

(d) Using the results from part (c) and the range rule of thumb, identify the range of usual value.

#### Solution

The study involve random selection of 12 companies. Given that $n = 12$ and $p = 0.80$. Let $X$ denote the number of companies test for substance abuse out of 12.

$X\sim B(12, 0.80)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &=\binom{12}{x}(0.80)^x(1-0.80)^{12-x},\\ &\quad x=0,1,\cdots, 12 \end{aligned}

(a) The probability that exactly 7 of the 12 companies test for substance abuse is

 \begin{aligned} P(X = 7) &= \binom{12}{7}(0.8)^7*(0.2)^5 \\ &=0.0532 \end{aligned}

(b) The probability that at least half of the companies test for substance abuse.

 \begin{aligned} P(7\leq X \leq 12) & = \bigg(\binom{12}{7}(0.8)^7*(0.2)^{5} +\binom{12}{8}(0.8)^8*(0.2)^{4}\\ &\quad +\cdots + \binom{12}{12}(0.8)^{12}*(0.2)^{0} \bigg)\\ &=\bigg(0.0532+ 0.1329+ 0.2362\\ &\quad +0.2835+0.2062 + 0.0687\bigg)\\ &= \big(0.9806\big) \end{aligned}

(c) For a group of 12 companies, find the mean and standard deviation for the number (among 12) that test for substance abuse.

 $$\text{ mean = } np = 12*0.8 = 9.6$$

 $$\text{sd = } \sqrt{np(1-p)}= \sqrt{12*0.8*0.2} = 1.3856$$

(d) Using the results from part (c) and the range rule of thumb, identify the range of usual value.
The range of usual values is $(mean- 2*sd, mean+2sd)$. Thus we get

 $$(9.6 -2*1.3856, 9.6+2*1.3856)$$

 $$(6.8288, 12.3712)$$

The values are unusual if they lie outside these limits.