Question

Table # 1 presents the probabilistic distribution of the clients that enter Wendy’s in a period of 15 minutes.

x	P(x)
5	.01
6	.02
7	.03
8	.05
9	.08
10	.09
11	.11
12	.13
13	.12
14	.10
15	.08
16	.06
17	.05
18	.03
19	.02
20	.01
21	.01

Find the expected value and standard deviation of this probability distribution. Find the 68% intervals. 95% and 99% (empirical rule). (Use Excel to answer this question). At the end you must explain in your words what these results mean

If it were a Poisson distribution, find:
What is the probability that exactly 16 customers will enter the fast food restaurant in the next 15 minutes?
What is the probability that 10 or fewer clients will enter in the next 15 minutes?
What is the probability that they enter between 5 <= x <= 15?

Answer 1

Solution

Back-up Theory

If a discrete random variable, X, has probability function, p(x), x = x1, x2, …., xn, then

Expected value = Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values ….........…. (1)

E(X²) = Σ{x².p(x)} summed over all possible values of x………………………………................................…..(2)

Variance of X = Var(X) = σ2 = E[{X – E(X)}²] = E(X²) – {E(X)}²…………..................................…...…………..(3)

Standard Deviation of X = SD(X) = σ = sq.rt of Var(X) ……………..............................……..……....…………..(4)

Empirical rule

Mean ± 1 Standard Deviation holds 68% of the observations; ………….....................................…………….(5a)

Mean ± 2 Standard Deviations holds 95% of the observations ……………..............................................….(5b)

and Mean ± 3 Standard Deviations holds 99.7% of the observations. ….............................................……..(5c)

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ……........................................……..(6)

This probability can also be obtained by using Excel Function, Statistical, POISSON ..........................…. (6a)

Now to work out the solution,

Let X = number of clients that enter Wendy’s in a period of 15 minutes

Part (a)

Vide (1),

Expected value of the probability distribution

= 12.47   Answer 1

Vide (4),

Standard deviation of the probability distribution

= 3.27   Answer 2

Details of calculations

x	p(x)	x.p(x)	x2.p(x)
5	0.01	0.05	0.25
6	0.02	0.12	0.72
7	0.03	0.21	1.47
8	0.05	0.40	3.20
9	0.08	0.72	6.48
10	0.09	0.90	9.00
11	0.11	1.21	13.31
12	0.13	1.56	18.72
13	0.12	1.56	20.28
14	0.1	1.40	19.60
15	0.08	1.20	18.00
16	0.06	0.96	15.36
17	0.05	0.85	14.45
18	0.03	0.54	9.72
19	0.02	0.38	7.22
20	0.01	0.20	4.00
21	0.01	0.21	4.41
Total	1	12.47	166.19
E(X)	12.47
E(X^2)	166.19
Var(X)	10.6891
SD(X)	3.27

Part (b) (empirical rule)

Vide (5a),

68% interval = 12.47 ± 3.27

= [9.20, 15.74]. Answer 3

Vide (5b),

95% interval = 12.47 ± 6.54

= [5.87, 19.01]. Answer 4

Vide (5c),

99% interval = 12.47 ± 6.54

= [2.56, 22.28] Answer 5

Part (c) Poisson distribution

Here, we have X ~ Poisson (12.47)
Probability that exactly 16 customers will enter the fast food restaurant in the next 15 minutes

= P(X = 16)

= 0.0627 [vide (6) and (6a)] Answer 6

Probability that 10 or fewer clients will enter in the next 15 minutes

= P(X ≤ 10)

= 0.3 [vide (6) and (6a)] Answer 7

Probability 5 ≤ x ≤ 15

P(X ≤ 15) - P(X ≤ 5)

= 0.8085 – 0.0151 [vide (6) and (6a)]

= 0.7934 Answer 8

DONE