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

Answer:

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

Step-by-step 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

$\sigma = \sqrt{p (1-p) n}$

$\sigma = \sqrt{0.20(1-0.20) 75}$

$\sigma = \sqrt{0.20*0.80* 75}$

$\sigma = \sqrt{12}$

$\sigma = 3.464$

Now ; if 14 or more people drank coffee ; then

$P(X \geq 14) = P(\dfrac{X-\mu }{\sigma} \leq \dfrac{X-\mu}{\sigma})$

$P(X \geq 14) =P(\dfrac{14-\mu }{\sigma} \leq \dfrac{14-15}{3.464})$

$P(X \geq 14) = P(Z \leq \dfrac{-1}{3.464})$

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

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