Question

A nationwide survey of 17,000 college seniors by
the University of Michigan revealed that almost 70%
disapprove of daily pot smoking. If 18 of these seniors
are selected at random and asked their opinion, what
is the probability that more than 9 but fewer than 14
disapprove of smoking pot daily?

Answer 1

Answer on Question #72630, Math / Statistics and Probability

A nationwide survey of 17,000 college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. If 18 of these seniors are selected at random and asked their opinion, what is the probability that more than 9 but fewer than 14 disapprove of smoking pot daily?

Solution

Let $N$ be the total number of senior students: $N = 17000$.

Let $K$ be the number of senior students disapproving of daily pot smoking: $K = 17000 \cdot 0.70 = 11900$.

Let $n$ be the number of senior students selected as sample: $n = 18$.

Let $X$ be the random variable denotes the number of senior students disapproving of daily pot smoking among the sample of $n$ students.

That is $X = 0,1,2,\ldots,18$.

Use the binomial distribution. We want to know $P(10 \leq X \leq 13)$.

From cumulative binomial tables

$$\begin{array}{l} P(10 \leq X \leq 13) = P(X \leq 13) - P(X \leq 9) = \\ = \sum_{x=0}^{13} b(x; 18, 0.7) - \sum_{x=0}^{9} b(x; 18, 0.7) \approx 0.667345 - 0.059586 \approx 0.60776 \\ \end{array}$$

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