Answer:
i) The sketch of the area under the normal distribution curve is attached to this solution of the question.
ii) The minimum number of seconds we could expect the longest 20% of customers to wait in line = 162 seconds.
Explanation:
This is is a normal distribution problem with
Mean = μ = 138 seconds
Standard deviation = σ = 29 seconds
i) Which of the following illustrates the shaded area under the normal distribution for the top 20%?
We first obtain the z-score that corresponds to the lower limit of the top 20% of the distribution of waiting times.
Let that z-score be z'
P(z > z') = 0.20
P(z > z') = 1 - P(z ≤ z') = 0.20
P(z ≤ z') = 1 - 0.20 = 0.80
P(z ≤ z') = 0.80
So, checking the normal distribution table,
z' = 0.842
we can then go ahead and obtain the waiting time that corresponds to this z-score.
Let the waiting time that corresponds to this z-score be x'
z' = (x' - μ)/σ
0.842 = (x' - 138)/29
x' = 162.42 seconds
Since, the options for the shaded area under the normal curve isn't presented with this question, the graph of the shaded area under the normal curve that corresponds to the top 20% waiting times is attached to this solution.
ii) What is the minimum number of seconds we could expect the longest 20% of customers to wait in line?
The minimum number of seconds we could expect the longest 20% of customers to wait in line corresponds to the lower limit of the top 20% waiting times obtained in (i) above.
The minimum number of seconds we could expect the longest 20% of customers to wait in line = 162.42 seconds = 162 seconds to the nearest second
Hope this Helps!!!