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 [tex]\sim[/tex] N (n, p)
X Â [tex]\sim[/tex] N (75, 0.20)
[tex]\mu = np[/tex] = 75 × 0.20
[tex]\mu =[/tex] 15
[tex]\sigma = \sqrt{p (1-p) n}[/tex]
[tex]\sigma = \sqrt{0.20(1-0.20) 75}[/tex]
[tex]\sigma = \sqrt{0.20*0.80* 75}[/tex]
[tex]\sigma = \sqrt{12}[/tex]
[tex]\sigma = 3.464[/tex]
Now ; if 14 or more people drank coffee ; then
[tex]P(X \geq 14) = P(\dfrac{X-\mu }{\sigma} \leq \dfrac{X-\mu}{\sigma})[/tex]
[tex]P(X \geq 14) =P(\dfrac{14-\mu }{\sigma} \leq \dfrac{14-15}{3.464})[/tex]
[tex]P(X \geq 14) = P(Z \leq \dfrac{-1}{3.464})[/tex]
[tex]P(X \geq 14) = P(Z \leq -0.28868)[/tex]
From the standard normal z tables; (-0.288)
[tex]P(X \geq 14) = P(Z \leq 0.38667)[/tex]
[tex]P(X \geq 14) = 1 - 0.38667[/tex]
[tex]P(X \geq 14) = 0.61333[/tex]
the probability that out of these 75 people, 14 or more drink coffee is 0.6133
