Answer:
(a) less than 10 minutes
= 0.5
(b) between 5 and 10 minutes
= 0.5
Step-by-step explanation:
We solve the above question using z score formula. We given a random number of samples, z score formula :
z-score is z = (x-μ)/ Standard error where
x is the raw score
μ is the population mean
Standard error : σ/√n
σ is the population standard deviation
n = number of samples
(a) less than 10 minutes
x = 10 μ = 10, σ = 2 n = 50
z = 10 - 10/2/√50
z = 0 / 0.2828427125
z = 0
Using the z table to find the probability
P(z ≤ 0) = P(z < 0) = P(x = 10)
= 0.5
Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5
(b) between 5 and 10 minutes
i) For 5 minutes
x = 5 μ = 10, σ = 2 n = 50
z = 5 - 10/2/√50
z = -5 / 0.2828427125
= -17.67767
P-value from Z-Table:
P(x<5) = 0
Using the z table to find the probability
P(z ≤ 0) = P(z = -17.67767) = P(x = 5)
= 0
ii) For 10 minutes
x = 10 μ = 10, σ = 2 n = 50
z = 10 - 10/2/√50
z = 0 / 0.2828427125
z = 0
Using the z table to find the probability
P(z ≤ 0) = P(z < 0) = P(x = 10)
= 0.5
Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is
P(x = 10) - P(x = 5)
= 0.5 - 0
= 0.5
