Question

Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 9 min and standard deviation 2 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?

Answer 1

When the sample size is
`n_1 = 5` ,

`P( X < 11) = 0.9875`

When the sample size is
`n_2 = 6` ,

`P( X < 11) = 0.993`

Explanation:

From the question we are told that

The mean is
`\mu = 9 \ min`

The standard deviation is
`\sigma = 2 \ min`

The first sample size is
`n_1 = 5`

The second sample size is
`n_2 = 6`

Generally the standard error of the mean is mathematically represented as

`\sigma_(x) = (\sigma)/( √(n) )`

=>
`\sigma_(x) = (2)/( √(5) )`

When the sample size is
`n_1 = 5` ,

Generally the standard error of the mean is mathematically represented as

`\sigma_(x) = (\sigma)/( √(n) )`

=>
`\sigma_(x) = (2)/( √(5) )`

=>
`\sigma_(x) = 0.894`

Generally the probability that the sample average amount of time taken on each day is at most 11 min is mathematically represented as

`P( X < 11) = P( (X - \mu )/(\sigma ) < (11 - 9 )/(0.8944 ) )`

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

`P( X < 11) = P( Z < 2.24 )`

From the z table the area under the normal curve to the left corresponding to 2.24 is

`P( Z < 2.24 ) = 0.9875`

=>
`P( X < 11) = 0.9875`

When the sample size is
`n_2 = 6` ,

Generally the standard error of the mean is mathematically represented as

`\sigma_(x) = (\sigma)/( √(n) )`

=>
`\sigma_(x) = (2)/( √(6) )`

=>
`\sigma_(x) = 0.816`

Generally the probability that the sample average amount of time taken on each day is at most 11 min is mathematically represented as

`P( X < 11) = P( (X - \mu )/(\sigma ) < (11 - 9 )/(0.816 ) )`

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

`P( X < 11) = P( Z < 2.45 )`

From the z table the area under the normal curve to the left corresponding to 2.24 is

`P( Z < 2.45 ) = 0.993`

=>
`P( X < 11) = 0.993`