Question

A population of values has a normal distribution with μ = 231.5 and σ = 53.2 . If a random sample of size n = 19 is selected

, Find the probability that a single randomly selected value is greater than 241.3. Round your answer to four decimals. P(X > 241.3) =

Find the probability that a sample of size n=19n=19 is randomly selected with a mean greater than 241.3. Round your answer to four decimals.
P(M > 241.3) =

Answer 1

Solution :

Given ,

mean = = 231.5

standard deviation = = 53.2

(A)P(x >241.3 ) = 1 - P(x<241.3 )

= 1 - P[(x -) / < (241.3-231.5) /53.2 ]

= 1 - P(z <0.18 )

Using z table

= 1 - 0.5714

= 0.4286

probability= 0.4286

(B)

n = 19

m = 231.5

m = / n = 53.2 / 19 =12.2049

P(M > 241.3) = 1 - P(M <241.3 )

= 1 - P[(M - m ) / m < (241.3 -231.5) /12.2049 ]

= 1 - P(z < 0.80)

Using z table

= 1 - 0.7881

= 0.2119

probability= 0.2119