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 probabilit…

Question

asked Jan 13, 2021 5.8k views

1 Answer

Mike Willis · Answer 1 · 2021-01-18T14:22:59+0000

Answer:

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