CoBE STAT-1
INDIVIDUAL ASSIGNMENT
GETAHUN ASSAB
UGR/7399/16
SUBMITTED TO Dr. MEBA
Exercise Questions
1
1. The average salary of the president and four vice presidents of
a company is $50,000. If vice
presidents’ average salary is $45,000, what is
the president’s salary?
2. Find the mean, median, mode, variance and
standard deviation of the following sample data.
class 40-44 45-49 50-54 55-59 60-64 65-69 70-74
frequency 7 10 22 15 12 6 3
3. Compute the mean deviation of the following data.
Price/ 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
kg
Units
order 2 12 15 20 18 10 9 4
4. Two distributions A and B have mean 80 and 20
inches and standard deviations 10 and 15 inches
respectively. Which distribution has more unlike
elements?
5. Suppose that the following data show the ratings of
hard-shell jackets based on the breath ability, durability,
versatility, features, mobility, and weight of each jacket.
The ratings range from 0 (lowest) to 100 (highest).
42, 66, 67, 71, 78, 62, 61, 76, 71, 67,
61, 64, 61, 54, 83, 63, 68, 69, 81. 53
a. Compute the mean, median, and mode.
b. Compute the first and third quartiles.
c. Compute and interpret the 90th percentile.
ANSWERS FOR THE QUESTIONS
2
1.Let the salary of the president be P, and the salary of each of the four
vice presidents be V 1,V 2,V 3,V 4 their mean are
p+ V 1+ V 2+V 3+V 4
=50,000
5
P+V 1 +V 2 +V 3 +V 4 = 250,000……………………………..*
V 1+V 2+V 3+V 4
=4 5,000
4
V 1 +V 2 +V 3 +V 4 = 180,000…………………………….**
P+180,000=250,000
P=250,000-180,000
P=70,000 This is the salary of the president.
2.Look the the following table.
Class Class 2
f i ( xi − μ) frequency Less than fxi
interval mark( x i) cumulative
frequency
40-44 42 1,183 7 0 294
45-49 47 640 10 7 470
50-54 52 198 22 17 1,144
55-59 57 60 15 39 855
60-64 62 588 12 54 744
65-69 67 864 6 66 402
70-74 72 867 3 72 216
total 4,400 75 75 4,125
LCL= Lower class Limit
fc = frequency of the class
Cfb = cumulative frequency of before the median class
N = number of observation
3
4,125
∑ f i xi= 4,125 2
f i ( x i − μ ) = 4,400 ∑ f = 75 Mean = 75
Mean = 55
[ ]
N
− fbc
Median = LCL+ 2
i ,LCL = 50 ,i = 5
fc
[ ]
75
− 17
Median = 50 + 2
22
Median = 50 + [ 37.5 −17
22
5 ]
Median = 50 + 4.65
Median ≈ 54.65
D1
[
Mo = LCL + D 1+ D 2 i , ]
D1= The difference of the frequency between the frequency of the class
and it’s preceding class.
D2= The difference of the frequency between the frequency of the class
and it’s subsequent class.
12
[ ]
Mo = 50 + 12+7 5 ,Mo = 50 + 3.15
Mo≈ 53.16
2
f 1 ( x 1 − μ )2+ f 2 ( x 2 − μ )2+ ...+ f n ( x n − μ )2
2 2 4,400 2
δ = variance, δ = ,δ = ,δ =
f 1+ f 2 +...+f n 75
58.67
Standard deviation = δ
δ= √δ2 δ= √ 58.67 δ ≈ 7.66
3.
4
Price/kg Units order Class mark f i|( xi − μ )| fxi
20-29 2 24.5 68 49
30-39 12 34.5 288 414
40-49 15 44.5 210 667.5
50-59 20 54.5 80 1090
60-69 18 64.5 108 1161
70-79 10 74.5 160 745
80-89 9 84.5 234 760.5
90-99 4 94.5 144 378
Total 90 1292 5265
First we must calculate the mean.Then our mean is 58.5
∑ f i|x i − μ|
MD(Mean Deviation) = f + f + ...+ f
1 2 n
1292
MD = 90 MD≈ 14.36
4. To determine the unlike elements or the more variable first we need to
calculate coefficient variable of both distributions.
σ 10
CV = μ x100 ,CV of A= 80 x100 ,CV of A =12.5
σ 15
CV of B = μ x100 , CV of B = 20 x100 ,CV of B = 75
The CV of B is more than Cv of A.Then the greater CV or standard
deviation is the more variable or has unlike elements.Therefore B has
unlike elements.
5.a
42+ 66+67+71+78+ 62+ 61+ 76+71+67+61+ 64+61+54 +83+63+68+ 69+81+53
μ=
20
1,318
μ=
20
μ=¿ 65.9
5
Md: first arrange the data an increasing order.
42, 53, 54, 61, 61, 61, 62, 63, 64, 66, 67, 67, 68, 69, 71, 71, 76, 78, 81, 83
66+67
Then Md = 2 , Md = 66.5
Mo = 61 the most frequency value among the data.
( ) item Q is the quartile value(Q ,Q ,Q )
th
k ( N +1 )
b .Qk = k 1 2 3
4
( ) item
tℎ
1 (20+ 1 )
Q 1= Q1 = (5.25)tℎ item
4
X 5.25 = X 5 + 0.25( X 6 - X 5 )
= 61+ 0.25(61-61) ∴ Q1=61
= 61
tℎ
Q 3= 3 (20 +1 ) Q3= (15.25 ¿ ¿tℎ item
4 item
X 15.25 = X 15 + 0.25( X 16 - X 15 )
= 71 + 0.25(71 -71) ∴ Q3 = 71
= 71
tℎ
90 ( 20+1 )
c . P90 = 100
item P90 = 18.9
Interpretation: P90 = 18.9 is the 0.9 or 90% of ways of between the value
in position 18 and the value in position 19.Then the value of
P90 = 18.9 given as follow:
X 18.9 = X 18 +0.9( X 19 - X 18 )
= 78 + 0.9(81 -78)
= 78 + 2.7
6
= 80.7
∴ P90 = 80.7 This means 90% of the jacket ratings are less than or equal to 80.7.
