Question

A population of values has a normal distribution with μ = 128.5 and σ= 26.6 You intend to draw a random sample of size n=58

Find the probability that a single randomly selected value is greater than 123.3. P(X > 123.3) =

Find the probability that a sample of size n= 58 is randomly selected with a mean greater than 123.3. P(M > 123.3) =

Answer 1

Solution :

Given that,

mean = = 128.5

standard deviation = = 26.6

n = 58

= 128.5

= / n = 26.6 / 58 = 3.4928

P( > 123.3) = 1 - P( < 1213.3 )

= 1 - P[( - ) / < ( 123.3 - 128.5) / 3.4928 ]

= 1 - P(z < -1.489)

Using z table,    

= 1 - 0.0682

= 0.9318

Probability = 0.9318

( b )

n = 58

M = 128.5

M = / n = 26.6 / 58 = 3.4928

P(M  > 123.3 ) = 1 - P( M < 123.3 )

= 1 - P[(M - M ) / M < ( 123.3 - 128.5) / 3.4928 ]

= 1 - P(z < -1.489)

Using z table,    

= 1 - 0.0682

= 0.9318

Probability = 0.9318