Question

A selective university advertises that 96% of its bachelor’s degree graduates have, on graduation day, a professional job offer or acceptance in a graduate degree program in their major area of study. In a sample of 227 recent graduates this was true of 209 of them. The probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is about:________

a. 0.015
b. 0.001
c. 0.131
d, 0.084

Answer 1

The probability is
`P( p < 0.9207) = 0.0012556`

Explanation:

From the question we are told

The population proportion is
`p = 0.96`

The sample size is
`n = 227`

The number of graduate who had job is k = 209

Generally given that the sample size is large enough (i.e n > 30) then the mean of this sampling distribution is

`\mu_x = p = 0.96`

Generally the standard deviation of this sampling distribution is

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

=>
`\sigma = \sqrt{(0.96 (1 - 0.96 ))/(227) }`

=>
`\sigma = 0.0130`

Generally the sample proportion is mathematically represented as

`\^ p = (k)/(n)`

=>
`\^ p = (209)/(227)`

=>
`\^ p = 0.9207`

Generally probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is mathematically represented as

`P( p < 0.9207) = P( (\^ p - p )/(\sigma ) < (0.9207 - 0.96)/(0.0130 ) )`

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

`P( p < 0.9207) = P(Z< -3.022 )`

From the z table the area under the normal curve to the left corresponding to -3.022 is

`P(Z< -3.022 ) = 0.0012556`

=>
`P( p < 0.9207) = 0.0012556`