Question

A local statistician is interested in the proportion of high school students that drink coffee. Suppose that 20% of all high school students drink coffee.

What is the probability that out of these 75 people, 14 or more drink coffee?

Answer 1

the probability that out of these 75 people, 14 or more drink coffee is 0.6133

Explanation:

Given that:

sample size n = 75

proportion of high school students that drink coffee p = 20% = 0.20

The proportion of the students that did not drink coffee = 1 - p

Let X be the random variable that follows a normal distribution

X
`\sim` N (n, p)

X
`\sim` N (75, 0.20)

`\mu = np` = 75 × 0.20

`\mu =` 15

`\sigma = √(p (1-p) n)`

`\sigma = √(0.20(1-0.20) 75)`

`\sigma = √(0.20*0.80* 75)`

`\sigma = √(12)`

`\sigma = 3.464`

Now ; if 14 or more people drank coffee ; then

`P(X \geq 14) = P((X-\mu )/(\sigma) \leq (X-\mu)/(\sigma))`

`P(X \geq 14) =P((14-\mu )/(\sigma) \leq (14-15)/(3.464))`

`P(X \geq 14) = P(Z \leq (-1)/(3.464))`

`P(X \geq 14) = P(Z \leq -0.28868)`

From the standard normal z tables; (-0.288)

`P(X \geq 14) = P(Z \leq 0.38667)`

`P(X \geq 14) = 1 - 0.38667`

`P(X \geq 14) = 0.61333`

the probability that out of these 75 people, 14 or more drink coffee is 0.6133