Question

The time needed to complete a job is normally distributed with a mean of 1 hour and standard deviation of 10 minutes. Probability to complete the job in hone hour or more, but less than 70 minutes is:

Answer 1

The value is
`P(60 < X < 70) = 0.34134`

Explanation:

From the question we are told that

The mean is
`\mu = 1 \ hour = 60 \ minutes`

The standard deviation is
`\sigma = 10 \ minutes`

Generally the 1 hour is equivalent to 60 minutes

Generally the probability to complete the job in one hour or more, but less than 70 minutes is mathematically represented as

`P(60 < X < 70) = P((60 - 60 )/(10 ) < (X - \mu)/(\sigma ) < (70 - 60 )/(10 ) )`

=>
`P(60 < X < 70) = P(0 < (X - \mu)/(\sigma ) < 1 )`

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

=>
`P(60 < X < 70) = P(0 <Z < 1 )`

=>
`P(60 < X < 70) = P(Z < 1) - P(Z < 0 )`

From the z table (Z < 1) and (Z < 0 ) is

`P(Z < 1) = 0.84134`

and

`P(Z < 0 ) = 0.5`

=>
`P(60 < X < 70) = 0.84134 - 0.5`

=>
`P(60 < X < 70) = 0.34134`