Question

A population of values has a normal distribution with μ = 96.2 μ = 96.2 and σ = 15 σ = 15 . You intend to draw a random sample of size n = 92 n = 92 .

Find the probability that a single randomly selected value is between 94 and 96.5. P(94 < X < 96.5) =

Find the probability that a sample of size n = 92 n = 92 is randomly selected with a mean between 94 and 96.5. P(94 < M < 96.5) =

Answer 1

Solution :

Given that ,

mean = = 96.2

standard deviation = = 15

P(94< x <96.5 ) = P[(94 - 96.2) / 15< (x - ) / < (96.5 - 96.2 ) /15 )]

= P( -0.15< Z <0.02 )

= P(Z <0.02 ) - P(Z < -0.15)

Using z table   

= 0.5080 - 0.4404

probability= 0.0676

b.

Solution :

Given that ,

mean = = 96.2

standard deviation = = 15

n = 92

sample distribution of sample mean is ,

=

= 96.2

sampling distribution of standard deviation

=  / n = 15/ 92=1.56

= 1.56

P(94 < M < 96.5) = = P[(94 - 96.2) /1.56 < ( - ) /   < (96.5 - 96.2) / 1.56)]

= P( -1.41< Z <0.19 )

= P(Z < 0.19) - P(Z <-1.41 )

Using z table

=0.5753 - 0.0793

=0.4960

  probability= 0.4960