Question

A survey of all medium- and large-sized corporations showed that of them offer retirement plans to their employees. Let be the proportion in a random sample of such corporations that offer retirement plans to their employees. Find the probability that the value of will be greater than .

Answer 1

Complete Question

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Answer:

a

`P(0.51 <  \^ p <  0.61 ) =  0.2587`

b

`P(\^ p > 0.71 ) = 0.15165`

Explanation:

From the question we are told that

The population proportion is
`p = 0.64`

The sample size is n = 50

Generally the mean of this sampling distribution is

`\mu_(x) = p = 0.64`

Generally the standard deviation is mathematically represented as

`\sigma  =  \sqrt{ (p(1- p ))/(n ) }`

=>
`\sigma  =  \sqrt{ (0.64(1- 0.64 ))/( 50) }`

=>
`\sigma  =  0.068`

Generally the probability that the value of
`\^ p` will be between 0.54 and 0.61 is mathematically represented as

`P(0.51 < \^ p < 0.61 ) = P( (0.5 4 - \mu_(x))/(\sigma ) < (\^ p- \mu_(x))/(\sigma ) < (0.6 1 - \mu_(x))/(\sigma ) )`

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

`P(0.51 <  \^ p <  0.61 ) =  P( (0.5 4  - 0.64)/(0.068 ) < Z < (0.6 1  - 0.64)/(0.068) )`

`P(0.51 <  \^ p <  0.61 ) =  P(-1.470 < Z < -0.4412 )`

=>
`P(0.51 <  \^ p <  0.61 ) =  P( Z < -0.4412 ) - P(Z <  -1.470 )`

From the z table the probabilities of ( Z < -0.4412 ) and (Z < -1.912 ) is

`P ( Z < -0.4412 )  = 0.32953`

and

`P(Z < -1.470 ) = 0.070781`

Generally

`P(0.51 <  \^ p <  0.61 ) =  0.32953 -0.070781`

`P(0.51 <  \^ p <  0.61 ) =  0.2587`

Generally the probability that the value of
`\^ p` will be greater than 0.71 is mathematically represented as

`P(\^ p > 0.71 ) = P( (\^ p - \mu_(x))/(\sigma) > (0.71 - 0.64 )/( 0.068 ) )`

=>
`P(\^ p > 0.71 ) = P( Z > 1.0294 )`

From the z table the probabilities of ( Z > 1.0294 )

`P( Z > 1.0294 ) = 0.15165`

So

`P(\^ p > 0.71 ) = 0.15165`