Question

In 2018 it was estimated that approximately 41% of the American population watches the Super Bowl yearly. Suppose a sample of 118 Americans is randomly selected. After verifying theconditions for the Central Limit Theorem are met, find the probability that the majority (more than 50%) watched the Super Bow.

Answer 1

We would test if np and n(1 - p) is at least 10

p = sample proportion

p = x/n

where

x = number of success

n = sample size

From the information given,

p = 41% = 41/100 = 0.41

n = 118

np = 48.38

n(1 - p) = 118(1 - 0.41) = 69.62

Both values are greater than or equal to 10. Thus, the distribution is approximately normal.

The mean of the distribution = p = 0.41

Standard deviation = √(p(1 - p)/n

By substituting the values,

standard deviation = √(0.41(1 - 0.41)/118 = √0.00205 = 0.045

We want to find P(p ≥ 0.5)

We would standardize 0.5 into a z score by applying the formula,

z = (x - mean)/standard deviation

z = (0.5 - 0.41)/0.045 = 2

From the normal distribution table, the probability value corresponding to a z score of 2 is 0.9773

P(p ≥ 0.5) = 1 - 0.9773

P(p ≥ 0.5) = 0.023

the probability that the majority (more than 50%) watched the Super Bow is 0.023