Question

A childcare centre adopts a strict late pickup policy. Parents who picked up their children late will be charged $1 per minute (assuming no rounding here, e.g., a late of 1.235 minutes will be charged $1.235). Bob is a parent who is always late, and each day the amount of time in minutes he is late follows an exponential distribution with mean 1. Further assume that the time he is late is always independent from day to day. His child will be in childcare for 256 days this year.

1) [4 marks] What is the expected late fees Bob will pay in total this year? What is the standard deviation of the total fees?

2) [6 marks] Estimate the probability that he will pay more than $260 in late fees. When computing the probabilities related to normal distributions, use the Standard Normal Curve Areas table discussed in cla

Answer 1

Solution

Back-up Theory

If X ~ Exponential(β), where β = (average inter-event time), E(X) = SD(X) = β ........………….………(1)

If X = ΣXi, such that Xi ~ Exp (βi) and Xi’s are independent, then X ~ Exp(β), where β = Σβi .......... (2)

If mean and standard deviation of X are µ and σ, then for a constant c, mean and standard

deviation of cX are cµ and cσ .............................................................................................................(3)

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .……….....................………...…(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables.......... (4a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ........................................…(4b)

Now to work out the solution,

Part (1)

Given, ‘each day the amount of time in minutes he is late follows an exponential

distribution with mean 1. Further assume that the time he is late is always independent

from day to day. His child will be in childcare for 256 days this year.’ implies vide (2),

Y = amount of time in minutes he is late in a year follows Exp(265). ……................................................…………….. (5)

Given late fee is $1 per minute,

Vide (1) and (5),

Expected late fees Bob will pay in total this year = $265 Answer 1

Standard deviation of the total fees = $265 Answer 2

Part (b)

Assuming Y ~ N(265, 265),

probability that he will pay more than $260 in late fees

= P(Y > 260)

= P[Z > {(260 - 265)/265}] [vide (4)]

= P(Z > - 0.0189)

= 0.5066 [vide (4a)] Answer 3

DONE