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Questions with Solution - Business Statistics I | MSC 287, Exams of Business Statistics

University of Alabama - HuntsvilleBusiness Statistics

Material Type: Exam; Class: MSC 297 Business Statistics I Section 3; Subject: Management Science; University: University of Alabama - Huntsville; Term: Fall 2004;

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

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Module 3.08: Old Exam Questions, Chapter 3 MSC 287 Dr. Stafford Revised 08/11/04 Page 3 of 2 pages
Key to Old Examination Questions, Chapter 3
Question 1
Data: 10, 5, 8, 9, 2, 12, 5, 7, 5, 8 n = 10 n/2 = 5
Sorted Data: 2, 5, 5, 5, 7, 8, 8, 9, 10, 12 3X = 71 3X2 = 581
Part I
10 1 Sample size: 10, count them.
7.1 2 Mean:
XXn
===
โˆ‘
//71 10 7.1
5 3 Mode: Using OET, it is 5 with three occurrences.
7.5 4 Median: Since n/2 = 5, Median =
XX
56
2
78
2
+=+=
7.5
8.5444 5 Variance:
sXX
n
n
2
2
22
1
581 71
10
9
=
โˆ’
โˆ’=
โˆ’
=
โˆ‘
โˆ‘
[]
8.5444
2.923 6 Standard Deviation:
ss
== =
2
85444. 2.923
10 7 Range: max - min = 12 - 2 = 10
4 8 IQR: P(25) = (.25)(10) = 2.5 Y 3; Q1 = P25 = X3 = 5
P(75) = (.75)(10) = 7.5 Y 8; Q3 = P75 = X8 = 9 IQR = Q3 - Q1 = 9 - 5 = 4
5 935
th percentile: P(35) = (.35)(10) = 3.5 Y 4; P35 = X4 = 5
Part II Grouped Data
Class fiMifi Mifi Mi2crfiri
1-3 12240.10.1
4-6 3 5 15 75 0.4 0.3
7-9 4 8 32 256 0.8 0.4
10-12 2 11 22 242 1.0 0.2
Total 10 71 577 1.0
7.1 10 Group mean:
XfM
n
g
ii
===
โˆ‘
71
10 7.1
8 11 Group Mode: Modal class is 7-9; midpoint of this class - - 8 - - is approx of mode.
7.8777 12 Group variance:
[]
sfM fM
n
n
g
ii ii
2
222
1
577 71
10
9
=
โˆ’
โˆ‘
โˆ’=
โˆ’
=
8.1
2.807 13 Std Deviation:
s
g
== =
s 8.1 2.846
g2
8 14 Group Median: Median class is 7-9; midpoint of this class approx. median.
5 15 35th percentile: crfi first $ 0.35 in class 4-6. Using class midpoint gives 5.
pf3

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Download Questions with Solution - Business Statistics I | MSC 287 and more Exams Business Statistics in PDF only on Docsity!

Key to Old Examination Questions, Chapter 3

Question 1

Data: 10, 5, 8, 9, 2, 12, 5, 7, 5, 8 n = 10 n/2 = 5 Sorted Data: 2, 5, 5, 5, 7, 8, 8, 9, 10, 12 3 X = 71 3 X 2 = 581 Part I

10 1 Sample size: 10 , count them.

7.1 2 Mean: X = (^) โˆ‘ X / n = 71 10/ = 7.

5 3 Mode: Using OET, it is 5 with three occurrences.

7.5 4 Median: Since n/2 = 5, Median =

X (^) 5 X 6

2

7 8

2

=

= 7.

8.5444 5 Variance: s

X

X

n n

2

2

2 2

1

581

71

10 9

=

โˆ’

โˆ’

=

โˆ’

โˆ‘ โˆ‘

[ ]

2.923 6 Standard Deviation: s = s = =

2 8 5444. 2.

10 7 Range: max - min = 12 - 2 = 10 4 8 IQR: P(25) = (.25)(10) = 2.5 Y 3; Q 1 = P 25 = X 3 = 5 P(75) = (.75)(10) = 7.5 Y 8; Q 3 = P 75 = X 8 = 9 IQR = Q 3 - Q 1 = 9 - 5 = 4 5 9 35 th^ percentile: P(35) = (.35)(10) = 3.5 Y 4; P 35 = X 4 = 5

Part II Grouped Data

Class f (^) i M (^) i f (^) i M (^) i f (^) i M (^) i^2 crf (^) i ri

1-3 1 2 2 4 0.1 0.

4-6 3 5 15 75 0.4 0.

Total 10 71 577 1.

7.1 10 Group mean: X

f M

n

g

i i = = =

โˆ‘ 71

10

8 11 Group Mode: Modal class is 7-9; midpoint of this class - - 8 - - is approx of mode.

7.8777 12 Group variance:

[ ]

s

f M

f M

n

g n

i i

i i 2

2

(^2 )

2.807 13 Std Deviation: s^ g =^ s^ g =^ 8.1^ = 2.

2

8 14 Group Median: Median class is 7-9; midpoint of this class approx. median. 5 15 35 th^ percentile: crf (^) i first $ 0.35 in class 4-6. Using class midpoint gives 5.

Question 2

First, sort the data into ascending order:

35.36 1 CV = ( s / X x ) 100 = ( .2 8284 / 8 ) x 100 =35 3554. {See Table below.}

6 2 For P 35 , i = (.35)(12) = [4.2]+^ = 5, and P 35 = X 5 = 6. {See Table above.} 2.167 3 See computation for MAD below. 8 4 See computation for mean below. 7.5 5 See computation for median below 6 6 Mode = most often occurring value = 6 (three of them). {See Table above.} 9 7 Range = MAX - MIN = 14 - 5 = 9. {See Table below.} 12 8 Sample size = number of observations in the sample = 12. {Count them with OET} 2.8284 9 See computation for s below. 8 10 See computation for s 2 below.

X i X i^2 X X

i โˆ’^

Median =

X 6 X 7

X = โˆ‘ X i / n = 96 12/ = 8

[ ]

s

x

x

n

n

2

2

(^2 )

GX = 96 GX 2 = 856

Question 3

5 1 Solving for n: 6.5(n-1) = 151 - 25n; 31.5n = 157.5; or n = 5.

O 3 If F

2 < 1, F > F

2 ; for example, if F

2 = 0.50, F = 0..

T 4 For the sample,

s

x x

n

2 i

Thus, MAD = variance for this sample. T 5 Frequencies of different classes in the sample could be the same value, or they could be different values. Thus {f 4 < f 3 } is a possible relationship.