Question

A population of values has a normal distribution with μ = 107.5 and σ = 9.7 . A random sample of size n = 133 is drawn.

a.) Find the probability that a single randomly selected value is between 106.4 and 110. Round your answer to four decimal places.

P ( 106.4 < X < 110 ) =

b.) Find the probability that a sample of size n = 133 is randomly selected with a mean between 106.4 and 110. Round your answer to four decimal places.

P ( 106.4 < M < 110 ) =

Answer 1

Solution:

a)

P(106.4 < x < 110) = P((106.4 - 107.5)/ 9.7) < (x - ) /  < (110 - 107.5) / 9.7) )

= P(-0.11 < z < 0.26)

= P(z < 0.26) - P(z < -0.11)

= 0.6026 - 0.4562 Using standard normal table,  

Probability = 0.1464

b)

n = 133

= 107.5

= / n = 9.7/ 133 = 0.8411

P(106.4 < M < 110) = P((106.4 - 107.4) /0.8411 <(M - ) / < (110 - 107.5) / 0.8411))

= P(-1.31 < Z < 2.97)

= P(Z < 2.97) - P(Z < -1.31) Using standard normal table,  

= 0.9985 - 0.0951

= 0.9034

Probability = 0.9034