Question

4. Suppose that your customer’s average order size is 50 units with a standard deviation of 11 units . From this large data set, select a sample size , n=24 and compute the sample average .

1. Find the probability that this sample average will be less than 52.

2. Find the probability that this sample average will be between 49 and 53

3. Find the probability that this sample average will be greater than 48.

Answer 1

a)

Here, μ = 50, σ = 2.2454 and x = 52. We need to compute P(X <= 52). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (52 - 50)/2.2454 = 0.89

Therefore,
P(X <= 52) = P(z <= (52 - 50)/2.2454)
= P(z <= 0.89)
= 0.8133

b)
Here, μ = 50, σ = 2.2454, x1 = 49 and x2 = 53. We need to compute P(49<= X <= 53). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (49 - 50)/2.2454 = -0.45
z2 = (53 - 50)/2.2454 = 1.34

Therefore, we get
P(49 <= X <= 53) = P((53 - 50)/2.2454) <= z <= (53 - 50)/2.2454)
= P(-0.45 <= z <= 1.34) = P(z <= 1.34) - P(z <= -0.45)
= 0.9099 - 0.3264
= 0.5835

c)
Here, μ = 50, σ = 2.2454 and x = 48. We need to compute P(X >= 48). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (48 - 50)/2.2454 = -0.89

Therefore,
P(X >= 48) = P(z <= (48 - 50)/2.2454)
= P(z >= -0.89)
= 1 - 0.1867 = 0.8133