Question

the customer service center in a large New York department store has determined that the amount of time spent with the customer about a complaint is normally distributed with the mean of 9.3 minutes and a standard deviation of 2.2 minutes what is the probability that 4 randomly chosen customer with the complaint the amount of time spent with solving the complaint will be as follows a less than 10 minutes be longer than 5 minutes see between 8 and 15 minutes

Answer 1

1)

Here, μ = 9.3, s = 2.2 , n = 4, σ = (s/sqrt(n))
= 2.2/sqrt(2)
σ = 1.1 and x = 10. We need to compute P(X <= 10). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (10 - 9.3)/1.1 = 0.64

Therefore,
P(X <= 10) = P(z <= (10 - 9.3)/1.1)
= P(z <= 0.64)
= 0.7389

2)

Here, μ = 9.3, σ = 1.1 and x = 10. We need to compute P(X >= 10). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (10 - 9.3)/1.1 = 0.64

Therefore,
P(X >= 10) = P(z <= (10 - 9.3)/1.1)
= P(z >= 0.64)
= 1 - 0.7389
= 0.2611

3)

Here, μ = 9.3, σ = 1.1, x1 = 8 and x2 = 15. We need to compute P(8<= X <= 15). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (8 - 9.3)/1.1 = -1.18
z2 = (15 - 9.3)/1.1 = 5.18

Therefore, we get
P(8 <= X <= 15) = P((15 - 9.3)/1.1) <= z <= (15 - 9.3)/1.1)
= P(-1.18 <= z <= 5.18)
= P(z <= 5.18) - P(z <= -1.18)
= 1 - 0.119
= 0.8810