Download Assignment I

Chapter 2:
2- Complete the following table:
Grades on Business
Frequency
Statistics Exam
A: 90-100
16
B: 80-89
36
C: 65-79
90
D: 50-64
30
F: Below 50
28
TOTAL
200
Relative Frequency
.08
.18
.45
.15
.14
1.00
For B: 80-89: 36 / 200 = .18
4- a)
Type of Response
Number
1-Letter and product replacement
2-Letter and good coupon
3-Letter and cents-off coupon
4-Letter and refund check
5-Letter, refund check and coupon
6-Letter only
7-No response
Total
42
36
29
23
12
76
67
286
Relative
Frequency
.147
.126
.101
.08
.042
.266
.234
1.00
Relative Frequency Plot of Responses
7
0.234
0.266
5
0.042
0.08
0.101
0.126
0.147
3
1
0
0.1
0.2
0.3
Relative Frequency
1
b) percentage of customers received a response to their letter of complaint?
286 – 67 = 219
219 / 286 * 100 = 76.57%
14- Consider the following sample data:
26
34
21
32
42
22
12
56
39
25
36
41
28
30
38
23
17
27
39
19
17
27
39
19
a) using the first digit, the stem possibilities are: 1, 2, 3, 4 and 5
b) stem-and-leaf display:
Stems
1
2
3
4
5
Leaves
2, 7, 9
1, 2, 3, 5, 6, 7, 8
0, 2, 4, 6, 8, 9, 9
1, 2
6
22- A sample of 20 measurements is shown here:
Class
1
2
3
4
5
6
34
12
21
26
32
39
Class Interval
10.5 – 15.5
15.5 – 20.5
20.5 – 25.5
25.5 – 30.5
30.5 – 35.5
35.5 – 40.5
Total
32
25
36
31
28
30
Class Frequency
1
2
4
5
4
4
20
38
23
Class Relative Frequency
.05
.1
.2
.25
.2
.2
1.00
Relative Frequency histogram
0.25
0.25
Relative
Frequency
26
22
0.2
0.2
0.15
0.2
0.1
0.1
0.05
0
0.2
0.05
10.5-15.5
15.5-20.5
20.5-25.5
25.5-30.5
Class Interval
2
30.5-35.5
35.5-40.5
Chapter 3:
2- Suppose a data set contains the observations 3, 8, 4, 5, 3, 4, 6. Find:
a.
b.
c.
d.
e.
 x  3  8  4  5  3  4  6  33
 x  (3)  (8)  (4)  (5)  (3)  (4)  (6)  175
 ( x  5)  (3  5)  (8  5)  (4  5)  (5  5)  (3  5)  (4  5)  (6  5)  20
 ( x  2)  (3  2)  (8  2)  (4  2)  (5  2)  (3  2)  (4  2)  (6  2)  71
( x)  (3  8  4  5  3  4  6)  33  1089
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
8- Find the mean and the median for the following sample of n=6 measurements: 7, 3, 4, 1, 5, 6
mean 
Sumofthexvalues
 x  7  3  4  1  5  6  4.333

numberofobservations
n
6
median 
45
 4 .5
2
Question #20
Set: 3, 9, 0, 7, 4
Range = 9 – 0 = 9
Variance = s2 = [∑x2 - (∑x)2 /n] / n – 1 = ∑x2 - n(xbar)2 / n – 1 = [155 – 4.6] / 4 = 37.6
Standard Deviation = square root [∑(x – xbar)2 / n – 1] = sr [5.76 + 19.36 + 21.16 + 5.76
+ 0.36] / 4 = sr(52.4 / 4) = sr(13.1) = 3.62
Problem 3.36
x
77
8.8
0
2.9
Mean(x)
58
11
-5
3
s
8
2
1.5
0.1
z score
2.375
-1.100
3.333
-1.000
3
From the values of Z Score it is clear that the value 0 lies the greatest distance above the
mean, and the value 8.8 lies the greatest distance below the mean.
Problem 3.60
Ticker Beta (x) x2
AL
1.489
2.217
BX
0.987
0.974
CMK
0.746
0.557
DOC
1.220
1.488
ECH
0.907
0.823
FNC
0.859
0.738
GS
0.722
0.521
HIA
1.736
3.014
ID
1.187
1.409
JCP
0.561
0.315
KO
0.548
0.300
LIT
1.502
2.256
MAT
1.662
2.762
NSM
2.014
4.056
PRD
1.358
1.844
REV
0.879
0.773
S
0.688
0.473
T
0.132
0.017
TL
1.137
1.293
UPJ
0.951
0.904
VO
1.317
1.734
WEN
1.731
2.996
WIN
0.196
0.038
XON
0.980
0.960
ZE
1.304
1.700
Total
26.813 34.165
a.
∑x =26.813
∑x2 =26.813
(∑x)2 =718.94
_
Mean xbar = 1.073
Variance, s2 = 0.225313
Standard deviation, s = 0.474671
Range : xbar - 2s = 0.123178, and
xbar + 2s = 2.021862
All 25 values lies between these ranges.
4
The Empirical rule states that close to 95% of observations will lie in this range
for symmetric distributions. It also states that the percentage will be near 100%
for highly skewed distributions.
In this case because all 25 values lie in this range we can say that the values are
symmetric and highly skewed.
b.
Sorting the 25 numbers in ascending order we can see that the 13th number. 0.987, is the
median.
This means that of the remaining 24 numbers, 50%, or 12 numbers are smaller than this
number and the other 12 numbers are greater than the mean.
0.132
0.196
0.548
0.561
0.688
0.722
0.746
0.859
0.879
0.907
0.951
0.980
0.987
1.137
1.187
1.220
1.304
1.317
1.358
1.489
1.502
1.662
1.731
1.736
2.014
5
c.
The 80th percentile
= 80 x (25 + 1)/100
= 21
Therefore the 21st number, 1.502, is in the 80th percentile.
d.
Z Score of S
= (x - xbar)/s
= (0.688 - 1.073)/ 0.474671
= -0.81008
The negative sign in the z score means that the observation lies to the left of the
mean. Because the absolute value
6