Applied Statistics
Question Paper
(Show your calculation steps clearly.
Correct you answer to 4 decimal places and report the
measurement unit when applicable)
Question 1
The household expenditure (HK$) of 12 sampled families with 4 members in Hong Kong in September 2019
were as follows:
8095
6745
4427
2192
4697
7549
4302
4209
3520
7621
2757
6441
(a) Find the sample mean and sample standard deviation of the data.
(b) Find the inter-quartile range of the data.
(c) Find the 15th percentile and 85th percentile of the data.
(d) Use the sampled data as reference, what is the probability that the household expenditure of a family with
4 members in Hong Kong in September 2019 was more than $5000?
(e) Use X to denote the household expenditure in a month and Y to denote the household saving in a month.
Assume Y = 6200 - 0.4X. Find the mean, standard deviation and the variance of variable Y in this
sample.
Question 2
The following is the probability distribution function of the number of members (X) in a Gym house in one
day.
x
200
300
400
500
P(X = x)
4k
3k
2k
k
(a) Find the value of k.
(b) Compute E(X) and σ(X).
(c) The daily profit Y (in $) of the Gym house depends on the number of people using its facilities.
assumed to follow the equation Y = 50X - 3000. Find E(Y), Var(Y) and σ(Y).
Y is
(d) The facility fee and the expense of the Gym house will be adjusted so that the adjusted daily profit W is
expressed as W = aX – 2800. With the assumption that the distribution of X will not be affected, find
the value of a so that the expected daily profit after the adjustment equals to $18200.
(e) Find the standard deviation of the daily profit after the adjustment.
(f) Another type of profit earned by the company is by selling healthy product. Suppose the summary
statistics of the daily profit (V) by selling healthy product is as follow:
E(V) = $3500
σ(V) = $1300
Combine the profit earned from the Gym house (after the adjustment) and by selling healthy product.
Find the expectation and standard deviation of total daily profit of the company.
Solution
Question 1
(a) Sample mean = $5212.9167
Sample standard deviation = $2003.6990
(b) The ordered array of the data:
2192 2757 3520 4209 4302 4427 4697 6441 6745 7549 7621 8095
25
25th percentile = $3864.5
(i = 12 (100) = 3, 25th percentile =
75th percentile = $7147
(i = 12 (100) = 9, 75th percentile =
75
3𝑟𝑑 𝑑𝑎𝑡𝑎+4𝑡ℎ 𝑑𝑎𝑡𝑎
2
9𝑡ℎ 𝑑𝑎𝑡𝑎+10𝑡ℎ 𝑑𝑎𝑡𝑎
2
IQR = 7147 – 3864.5 = $3282.5
15
(c) 15th percentile = $2757 (i = 12 (100) = 1.8↑2, 15th percentile = 2nd data)
85
85th percentile = $7621 (i = 12 (100) = 10.2↑11, 85th percentile = 11th data)
(d) P(expenditure > $5000) =
5
12
= 0.4167
(e) With Y = 6200 – 0.4X
E(Y) = 6200 – 0.4E(X) = $4114.8333
σ(Y) = |-0.4|σ(X) = $801.4796
Var(Y) = σ(Y)2 = 642369.5492
)
)
Question 2
(a) With total probability = 1, 4k + 3k + 2k + k = 1, k = 0.1
(b) E(X) = 200(0.4) + 300(0.3) + 400(0.2) + 400(0.1) = 300 people
Var(X) = 2002 (0.4) + 3002 (0.3) + 4002 (0.2) + 5002 (0.1) – (300)2 = 10,000
σ(X) = √10000 = 100 people
(c) Y = 50X – 3000
E(Y) = 50E(X) – 3000 = 50(300) – 3000 = $12,000
σ(Y) = 50σ(X) = 50(100) = $5,000
Var(Y) = σ(X)2 = 25000000
(d) W = aX – 2800
E(W) = aE(X) – 2800 = 18200
a=
18200+2800
300
= 70
(e) W = 70X – 2800
σ(W) = 70σ(X) = 70(100) = $7000
(f)
Use T to denote the total profit, T = W + V
E(T) = E(W) + E(V) = 18200 + 3500 = $21700
Var(T) = Var(W) + Var(V) = (7000)2 + (1300)2 = 50690000
σ(T) = √𝑉𝑎𝑟(𝑇) = $7119.6910