Question

According to a study published by a group of University
of Massachusetts sociologists, approximately
60% of the Valium users in the state of Massachusetts
first took Valium for psychological problems. Find the
probability that among the next 8 users from this state
who are interviewed,
(a) exactly 3 began taking Valium for psychological
problems;
(b) at least 5 began taking Valium for problems that
were not psychological

Answer 1

Answer on Question #72632, Math / Statistics and Probability

According to a study published by a group of University of Massachusetts sociologists, approximately 60% of the Valium users in the state of Massachusetts first took Valium for psychological problems. Find the probability that among the next 8 users from this state who are interviewed

(a) exactly 3 began taking Valium for psychological problems;

(b) at least 5 began taking Valium for problems that were not psychological.

Solution

Let $X$ be the random variable representing the number of Valium users who began taking Valium for psychological problems out of next 8 users interviewed. The probability of a success in each trial is $p = 0.6$. Trials are independent. Hence, $X$ has a binomial distribution with parameters $n = 8$ and $p = 0.6$

$$X \sim \operatorname{Bin}(n, p), \quad \text{where } n = 8 \text{ and } p = 0.6$$

The probability mass function $(p.m.f)$ is

$$P(X = x) = b(x; 8, 0.6), \text{where } x = 0, 1, 2, \dots, 8$$ $$P(X = x) = \binom{8}{x} (0.6)^x (1 - 0.6)^{8 - x} = \binom{8}{x} (0.6)^x (0.4)^{8 - x},$$

where $x = 0,1,2,\ldots,8$

(a) Exactly 3 began taking Valium for psychological problems

$$\begin{aligned} P(X = 3) &= \binom{8}{3} (0.6)^3 (0.4)^{8 - 3} = \frac{8!}{3! (8 - 3)!} (0.6)^3 (0.4)^5 = \\ &= \frac{8(7)(6)}{1(2)(3)} (0.6)^3 (0.4)^5 = 0.12386304 \end{aligned}$$

(b) At least 5 began taking Valium for problems that were not psychological

$$\begin{aligned} P(X \leq 3) &= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = \\ &= \binom{8}{0} (0.6)^0 (0.4)^{8 - 0} + \binom{8}{1} (0.6)^1 (0.4)^{8 - 1} + \binom{8}{2} (0.6)^2 (0.4)^{8 - 2} + \\ &+ 0.12386304 = (0.4)^8 + 8(0.6)(0.4)^7 + \frac{8!}{2! (8 - 2)!} (0.6)^2 (0.4)^6 + \\ &+ 0.12386304 = 0.1736704 \end{aligned}$$

Answer provided by AssignmentExpert.com