Question

A normal population has a mean of 64 and a standard deviation of 24. You select a random sample of 32. Use Appendix B.1 for the z-values. Compute the probability that the sample mean is: (Round the final answers to 4 decimal places.)

a. Greater than 67.

Probability

b. Less than 60.

Probability

c. Between 60 and 67.

Probability

Answer 1

a)
Here, μ = 64, σ = 4.2426 and x = 67. We need to compute P(X >= 67). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (67 - 64)/4.2426 = 0.71

Therefore,
P(X >= 67) = P(z <= (67 - 64)/4.2426)
= P(z >= 0.71)
= 1 - 0.7611 = 0.2389

b)

Here, μ = 64, σ = 4.2426 and x = 60. We need to compute P(X <= 60). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (60 - 64)/4.2426 = -0.94

Therefore,
P(X <= 60) = P(z <= (60 - 64)/4.2426)
= P(z <= -0.94)
= 0.1736

c)

Here, μ = 64, σ = 4.2426, x1 = 60 and x2 = 67. We need to compute P(60<= X <= 67). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (60 - 64)/4.2426 = -0.94
z2 = (67 - 64)/4.2426 = 0.71

Therefore, we get
P(60 <= X <= 67) = P((67 - 64)/4.2426) <= z <= (67 - 64)/4.2426)
= P(-0.94 <= z <= 0.71) = P(z <= 0.71) - P(z <= -0.94)
= 0.7611 - 0.1736
= 0.5875