Question

Given a normal population whose mean is 615 and whose standard deviation is 68, find each of the following: A. The probability that a random sample of 6 has a mean between 619 and 646. Probability = B. The probability that a random sample of 19 has a mean between 619 and 646. Probability = C. The probability that a random sample of 22 has a mean between 619 and 646. Probability =

Answer 1

Solution :

Given that,

mean = = 615

standard deviation = = 68

(A)

n = 6

= 615

= / n = 68 / 6

P(619 646) = P((619 - 615) / 68 / 6 ( - ) / (646 - 615) / 68 / 6))

= P(0.1441 Z 1.1167)

= P(Z 1.1167) - P(Z 0.1441) Using z table,

= 0.8679 - 0.5573 = 0.3106

Probability = 0.3106

(B)

n = 19

= 615

= / n = 68 / 19

P(619 646) = P((619 - 615) / 68 / 19 ( - ) / (646 - 615) / 68 / 19))

= P(0.2564 Z 1.9871)

= P(Z 1.9871) - P(Z 0.2564) Using z table,

= 0.9765 - 0.6012 = 0.3753

Probability = 0.3753

C)

n = 22

= 615

= / n = 68 / 22

P(619 646) = P((619 - 615) / 68 / 22 ( - ) / (646 - 615) / 68 / 22))

= P(0.2759 Z 2.1383)

= P(Z 2.1383) - P(Z 0.2759) Using z table,

= 0.9838 - 0.6087 = 0.3751

Probability = 0.3751