SCHOOL OF MANAGEMENT
SEMESTER 1, ACADEMIC SESSION 2020/2021
ATW 123 BUSINESS STATISTICS
TUTORIAL CHAPTER 5
GROUP 4 POCKY
GROUP MEMBERS:
No. Name Matric No.
1 Arianna Lam Kah Yee 143982
2 Au Chun Tuan 142677
3 Chia Jian Wei 142650
4 Chua Chuan Rou 141890
5 Gayathiri Ravindran 141253
5.7 (page 241)
a. Expected return for stock X and Y
μ = E ( X ) = ∑ xi P ( x= xi )
Expected Return for Stock X Expected Return for Stock Y
(-60) * 0.1 = -6 (-130) * 0.1= -13
(20) * 0.2 = 4 (60) * 0.2 = 12
(100) * 0.4 = 40 (150) * 0.4 = 60
(160) * 0.3 = 48 (200) *0.3 = 60
Sum ∑ $ 86 $ 119
b. Standard Deviation for stock X and Y.
σ = √∑¿ ¿
Variance for Stock X Variance for Stock Y
(−60−86)2 ×(0.1) (−130−119)2 ×(0.1)
= 2131.6 = 6200.01
(20−86)2 ×(0.2) (60−119)2 ×(0.2)
= 871.2 = 696.2
(100−86)2 ×(0.4) (150−119)2 ×(0.4)
= 78.4 = 384.4
(160−86)2 ×(0.3) (200−119)2 ×(0.3)
= 1642.8 = 1968.3
Sum ∑ σ2 = $ 4724 σ2 = $ 9248.91
σ = $ 68.7314 σ = $ 96.1713
c. I would invest in Stock X because it has the lower risk as the standard deviation for stock
X is smaller than the standard deviation for stock Y which is $ 68.7414 and $ 96.1713.
Although the expected return of stock Y is greater than the expected return of Stock X which
is $119 and $ 86 respectively but due to the lower risk, I would invest in Stock X.
5.17 (page 248)
a. P (X = 3 | n = 3, π=0.922 ¿
3!
= ( 0.922 )3 (1−0.922)3 −3
3! (3−3 ) !
= 1(0.7838) * (1)
= 0.7838
b. P (X = 0 | n = 3, π=0.922 ¿
3!
= ( 0.922 )0 (1−0.922)3−0
0 ! ( 3−0 ) !
= 1(1) * (0.0004746)
= 0.0004746
c. P (X ≥ 2 | n = 3, π=0.922 ¿
= P (X = 2) + P (X = 3)
3! 3!
= ( 0.922 )2 (1−0.922)3−2+ ( 0.922 )3 (1−0.922)3 −3
2! ( 3−2 ) ! 3! (3−3 ) !
= 3(0.8501) * (0.078) + 1(0.7838) * (1)
= 0.1989 + 0.7838
= 0.9827
d. μ = E(x) = nπ
= 3(0.922)
= 2.766
σ = √ nπ (1−π )
= √ 3(0.922)(1−0.922)
= √ 0.2157
= 0.4644
The mean number of orders filled correctly is 2.766, and the standard deviation is 0.4644.
The probability that all three orders are filled correct is 0.7838 or 78.38%. The probability
that none of the orders are filled correctly is 0.0004746 or 0.04746%. The probability that at
least two orders are filled correctly is 0.9827 or 98.27%.
e. All 3 orders will be filled correctly P( X =3)
Problem 5.16 3!
( 0.905 )3( 1−0. 905)3−3
3! (3−3 ) !
= 0.7412
Example 5.4 0.6562
Problem 5.17 0.7838
None of orders will be filled correctly P( X =0)
Problem 5.16 3!
( 0.9 05 )0 (1−0.905)3−0
0 ! ( 3−0 ) !
= 0.0008674
Example 5.4 0.0022
Problem 5.17 0.0004746
At least 2 of the 3 orders will be filled correctly P( X ≥ 2)
Problem 5.16 3! 3!
( 0.9 05 )2 (1−0.9 05)3−2+ ( 0.905 )3 (1−0.9 05)3−3
2! ( 3−2 ) ! 3! (3−3 ) !
= 0.2334 + 0.7412
= 0.9746
Example 5.4 0.2968
Problem 5.17 0.9827
Mean
Problem 5.16 3(0.905)
= 2.715
Example 5.4 2.607
Problem 5.17 2.766
Standard Deviation
Problem 5.16 √ 3(0.905)(1−0.905)
= 0.5079
Example 5.4 0.5844
Problem 5.17 0.4644
Comparing to the Wendy’s in Example 5.4 on page 247, the mean number of orders filled
correctly has increased from 2.607 to 2.766 but the standard deviation decreased from 0.5844
to 0.4644. Besides, the probability that all three orders are filled correct has increased from
65.62% to 78.38%. The probability that none of the orders are filled correctly is decreased
from 0.22% to 0.04746%. Lastly, the probability that at least two orders are filled correctly
has increased from 29.68% to 98.27%.
Besides, comparing to the Burger King’s in Problem 5.16, the mean number of orders filled
correctly has increased from 2.715 to 2.766 but the standard deviation decreased from 0.5079
to 0.4644. Besides, the probability that all three orders are filled correct has increased from
74.12% to 78.38%. The probability that none of the orders are filled correctly is decreased
from 0.086%% to 0.04746%. Lastly, the probability that at least two orders are filled
correctly has increased from 97.46% to 98.27%.
5.25 (page 251)
a. P (X = 2 | λ = 10)
e−10 102
=
2!
= 0.002270
b. P (X < 2 | λ = 10)
= P (X = 0) + P (X = 1)
e−10 100 e−10 101
= +
0! 1!
= 0.00004540 + 0.0004540
= 0.0004994
c. P (X = 2 | λ = 5)
e−5 52
=
2!
= 0.08422
5.29 (page 252)
a. The probability of receiving a phone call within a certain interval does not change over
different intervals.
a. P (X = 0 | λ = 0.8)
e−0.8 0.80
=
0!
= 0.4493
b. P (X ≥ 3 | λ = 0.8)
= 1 – P (X ≤ 2)
= 1 – [P (X = 0) + P (X = 1) + P (X = 2)]
e−0.8 0.80 e−0.8 0.81 e−0.8 0.82
=1–[ + + ]
0! 1! 2!
= 1 – [0.4493 + 0.3595 + 0.1438]
= 1 – 0.9526
= 0.0474
c. P (X ≤ 3 | λ = 0.8)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
e−0.8 0.80 e−0.8 0.81 e−0.8 0.82 e−0.8 0.83
= + + +¿
0! 1! 2! 3!
= 0.4493 + 0.3595 + 0.1438 + 0.0383
= 0.9909
P (X ≤ 4 | λ = 0.8)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
e−0.8 0.80 e−0.8 0.81 e−0.8 0.82 e−0.8 0.83 e−0.8 0.84
= + + +¿ +
0! 1! 2! 3! 4!
=0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.007669
=0.9986
P (X ≤ 5 | λ = 0.8)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)
e−0.8 0.80 e−0.8 0.81 e−0.8 0.82 e−0.8 0.83 e−0.8 0.84 e−0.8 0.85
= + + +¿ + +¿
0! 1! 2! 3! 4! 5!
= 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.007669 + 0.001227
= 0.9998
P (X ≤ 6 | λ = 0.8)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
e−0.8 0.80 e−0.8 0.81 e−0.8 0.82 e−0.8 0.83 e−0.8 0.84 e−0.8 0.85 e−0.8 0.86
= + + +¿ + +¿ +
0! 1! 2! 3! 4! 5! 6!
= 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.007669 + 0.001227 + 0.0001636
= 1.0000
Therefore, the maximum number of phone calls that will be received in one-minutes period
99.99% of the time is 5 because the probability is approximately to 99.99% which is 99.98%.
