Question

Suppose the length of time a person takes to use an ATM at the bank is normally distributed with mean of 110 seconds and standard deviation of 10 seconds. There are 4 people ahead of you in the queue waiting to use the machine. You are concerned about the total time (T) the 4 people ahead of you will take to use the machine.

(i) What is the mean value of T, the total time (in seconds) for the 4 people ahead of you to use the machine?

(ii) Assuming that the times for the 4 people are independent of each other, determine the standard deviation of T. (remember 20 x 20 = 400) (iii) Sketch the distribution for T, clearly labelling the important features on your sketch.

(iv) Use your sketch and the Empirical rule (0.68 within 1 standard deviation of the mean, 0.95 within 2 standard deviations of the mean, 0.997 within 3 standard deviations of the mean in a normal population) to find the probability that the total time T is less than 400 seconds.

Answer 1

(i)

Let X1, X2, X3 and X4 be the length of time 4 people ahead of you takes to use an ATM at the bank Then Xi ~ N(110, 10)

Total time, T = X1 + X2 + X3 + X4

Mean value of T = E[T] = E[X1 + X2 + X3 + X4] = E[X1] + E[X2] + E[X3] + E[X4] = 110 + 110 + 110 + 110 = 440 seconds

(ii)

Variance of T = Var(T) = E[X1 + X2 + X3 + X4] = Var[X1] + Var[X2] + Var[X3] + Var[X4] = 10² + 10² + 10² + 10² = 400

Standard deviation of T = = 20 seconds

(iii)

(iv)

Probability that the total time T is less than 400 seconds = P(T < 400)

Now,

Probability that the total time T is within 2 standard deviations of the mean = P(440 - 2 * 20 < T < 440 + 2 * 20) = 0.95

=> P(400 < T < 480) = 0.95

=> P(T < 400) + P(T > 480) = 1 - 0.95 = 0.05

Based on symmetry of Normal distribution,

P(T < 400) = 0.05/2 = 0.025