Question

The average age of CEOs is 56 years. Assume the variable is normally distributed, with a standard deviation of 4 years. Give numeric answers with 4 decimal places.

a) If one CEO is randomly selected, find the probability that he/she is older than 63. Blank 1

b) If one CEO is randomly selected, find the probability that his/her mean age is less than 57. Blank 2

c) If one CEO is randomly selected, find the probability that his/her age will be between 53 and 59. Blank 3

d) If 36 CEOs are randomly selected, find the probability that their mean age is between 53 and 59. Blank 4

e) Explain the reason the answers to c) and d) above are differen

Answer 1

a)

Here, μ = 56, σ = 4 and x = 63. We need to compute P(X >= 63). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (63 - 56)/4 = 1.75

Therefore,
P(X >= 63) = P(z <= (63 - 56)/4)
= P(z >= 1.75)
= 1 - 0.9599 = 0.0401

b)

Here, μ = 56, σ = 4 and x = 57. We need to compute P(X <= 57). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (57 - 56)/4 = 0.25

Therefore,
P(X <= 57) = P(z <= (57 - 56)/4)
= P(z <= 0.25)
= 0.5987

c)

Here, μ = 56, σ = 4, x1 = 53 and x2 = 59. We need to compute P(53<= X <= 59). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (53 - 56)/4 = -0.75
z2 = (59 - 56)/4 = 0.75

Therefore, we get
P(53 <= X <= 59) = P((59 - 56)/4) <= z <= (59 - 56)/4)
= P(-0.75 <= z <= 0.75) = P(z <= 0.75) - P(z <= -0.75)
= 0.7734 - 0.2266
= 0.5468

d)

Here, μ = 56, σ = 0.667, x1 = 53 and x2 = 59. We need to compute P(53<= X <= 59). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (53 - 56)/0.667 = -4.5
z2 = (59 - 56)/0.667 = 4.5

Therefore, we get
P(53 <= X <= 59) = P((59 - 56)/0.667) <= z <= (59 - 56)/0.667)
= P(-4.5 <= z <= 4.5) = P(z <= 4.5) - P(z <= -4.5)
= 1 - 0
= 1

e)

because sample sizes are different