Question

Research in March 2019 suggests that 40% of U.S. adults approve of way President Trump is running the country. We randomly sample 50 U.S. adults and find that 35% approve of way President Trump is running the country.

What is the probability that a random sample of 50 U.S. adults has less than 35% with this opinion? (Round standard error to 2 decimal places before calculating Z. Round Z to 2 decimal places before using the Normal Distribution Calculator.)

Answer 1

Probability that a random sample of 50 U.S. adults has less than 35% with this opinion is 0.22965.

Explanation:

We are given that Research in March 2019 suggests that 40% of U.S. adults approve of way President Trump is running the country. We randomly sample 50 U.S. adults and find that 35% approve of way President Trump is running the country.

Let
`\hat p` = sample proportion of U.S.adults politicians who approve of way President Trump is running the country.

The z-score probability distribution for sample proportion is given by;

Z =
`\frac{ \hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion

p = population proportion = 40%

n = sample of U.S. adults = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that a random sample of 50 U.S. adults has less than 35% with this opinion is given by = P(
`\hat p` < 0.35)

P(
`\hat p` < 0.51) = P(
`\frac{ \hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` <
`\frac{0.35-0.40}{\sqrt{(0.35(1-0.35))/(50) } }` ) = P(Z < -0.74) = 1 - P(Z
`\leq` 0.74)

= 1 - 0.77035 = 0.22965

Now, in the z table the P(Z
`\leq` x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.

Therefore, probability that a random sample of 50 U.S. adults has less than 35% with this opinion is 0.22965.