Question

A 2020 Pew Research poll found that 65% of U.S. adults believe that President Donald Trump was too slow to make major steps to stop the spread of the coronavirus. Let's assume that 0.65 is the population proportion. Suppose random samples of 100 U.S. adults are randomly selected, and the proportion of each sample is found.

What is the probability that a random sample of 100 adults have a sample proportion greater than 70%? Round the probability to 2 decimal places.

Answer 1

Solution :

Given that ,

p = 0.65

1 - p = 0.35

n = 100

$\mu _{\hat{p}}$ = p = 0.65  

$\sigma _{\hat{p}}$ = $\sqrt$ (p*(1-p))/n =  $\sqrt$(0.65*0.35)/100 = 0.04770

P($\hat{p}$ > 0.70) = 1 - P($\hat{p}$ < 0.70)

= 1 - P(($\hat{p}$ - $\mu _{\hat{p}}$ ) / $\sigma _{\hat{p}}$ < (0.70-0.65) /0.04770 )

= 1 - P(z < 1.048)

= 1 - 0.8527

= 0.1473

Probability = 0.15