80% of teenagers have access to some social media . If you do a random survey of 400 teenagers, what is the probability that between 300 and 340 of them will patronize it.

P(300 < p^ < 340) = 0.9876 or 98.76%

Step-by-step explanation:

We are given;

Population proportion; p = 80% = 0.8

Sample proportion; p1^ = 300/400 = 0.75 and p2^ = 340/400 = 0.85

Sample size: n = 400

z-score formula for sample proportion is;

z = (p^ - p)/(√(p(1 - p)/n)

Thus;

z1 = (0.75 - 0.8)/(√(0.8(1 - 0.8)/400)

z1 = -0.05/0.02

z1 = -2.5

z2 = (0.85 - 0.8)/(√(0.8(1 - 0.8)/400)

z2 = 0.05/0.02

z2 = 2.5

From online calculator of probability between 2 z-scores, we have;

P(-2.5 < z > 2.5) ≈ 0.9876

the correct answer is the second option