Question

Suppose that 50% of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than 0.585. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only .50? (Round the answer to four decimal places.)

Answer 1

The probability is
`P( p  >  0.585) =  0.044565`

Explanation:

From the question we are told that

The population proportion is p = 0.50

The sample size is n = 100

Gnerally the mean of the sampling distribution is

`\mu_x =  0.50`

Generally the standard deviation of the sampling distribution is

`\sigma  =  \sqrt{(p(1- p))/(n) }`

=>
`\sigma  =  \sqrt{(0.5 (1- 0.5))/(100) }`

=>
`\sigma  =0.05`

Generally the approximate probability that the cable company will keep the shopping channel

`P( p  >  0.585) =  P((p- \mu_(x))/(\sigma )  > (0.585 - 0.50)/(0.05) )`

Generally
`(p- \mu_(x))/(\sigma ) = Z (The  \  standardized \  value  \  of  p )`

=>
`P( p  >  0.585) =  P(Z > 1.7 )`

From the z-table the probability of (Z > 1.7 ) is

`P(Z > 1.7 ) = 0.044565`

So
`P( p  >  0.585) =  0.044565`