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.) Group of answer choices about 0.24 about 0.76 about 0.48

Answer 1

The probability is
`P( \^ p  <  0.35)=    0.2388`

Explanation:

From the question we are told that

The population proportion is p = 0.40

The sample size is n = 50

Generally the standard error is mathematically represented as

`\sigma_(\= p) =  \sqrt{(p *  (1- p))/(n) }`

=>
`\sigma_(\= p) =  \sqrt{(0.40 *  (1- 0.40))/(50) }`

=>
`\sigma_(\= p) = 0.07`

Generally the probability that a random sample of 50 U.S. Adults has less than 35% with this opinion is mathematically represented as

`P( \^ p  <  0.35) =  P((\^ p  -  p)/(\sigma_(\= p)) <  (0.35 - 0.40)/(0.07)  )`

Generally
`(\^ p  -  p)/(\sigma_(\= p)) =  Z (The \  standardized \ value\ of \  \^ p )`

=>
`P( \^ p  <  0.35)=P(Z  <  -0.71)`

From the z table

`P(Z  <  -0.71) =  0.2388`

So

`P( \^ p  <  0.35)=    0.2388`