Solution of assignment 2
2.36
157  132  109  145  125  139
 134.5
6
(157 – 134 .5) 2  (132 – 134 .5) 2  (139 – 134 .5) 2
s2 
= 276.7
6 –1
x
a.
s  276.7  16.63
x
2
i
Also
 x 
2
i
s2 
 157 2  132 2  109 2  145 2  125 2  139 2  109,925
 (157  132  109  145  125  139 ) 2  (807 ) 2  651,249
1 
n – 1 
x
2
i
–
1
n
 x    6 1– 1 109,925 – 16 (651,249)  276.7

2
i
b.
[ x  s]  [134.5  16.63]  [117.87,151.13]
[ x  2s]  [134.5  2(16.63)]  [101.24,167.76]
[ x  3s]  [134.5  3(16.63)]  [84.61,184.39]
c.
Yes, because $190 is not within the 99.73% interval [84.16,184.39]
157  134.5
= 1.353 It means that 157 is 1.353 standard deviation
16.63
above the mean.
d. z157=
z132 = –0.150
z109 = –1.533
z145 = 0.631
z125 = –0.571
z139 = 0.271
a. Histogram for Waiting Time data
2.38
25
Count
20
15
10
5
2.0
4.0
6.0
8.0
10.0
WaitTime
It is somewhat reasonable. Because it is approximate to a normal distribution
b.
[ x  s]  [5.46  2.475]  [2.985,7.935]
[ x  2s]  [5.46  2(2.475)]  [.51,10.41]
[ x  3s]  [5.46  3(2.475)]  [–1.965,12.885]
c.
Yes, because the upper limit of the 68.26% interval 7.935 is less than 8
minutes.
d.
66% fall into [ x  s], 96% fall into [ x  2s] , 100% fall into [ x  3s] .They
are approximate 68.26%,95.44%,99.73% respectively. So, the inference are
reasonably valid.
2.46
a. Box-and-whiskers display for the return on capital percentages
40
31
30
20
10
10
0
Re turn on Ca pita l (% )
b. The return percentages have a mean of 15.90, a median of 15.40. The
distribution is reasonably symmetric with one high outlier and one lower
outlier.
2.54
Bar chart for the poll’s results
De g re e o fF a m ilia r
60
50
F re q u e n c y
40
30
20
10
0
Ve ry Fa milia r
S ome wha t Fa milia r
Not too Fa milia r
De g re e o fF a m ilia r
Not a t a ll Fa milia r
Pie chart for the poll’s results
De g re e o fF a m ilia r
Ve ry Fa milia r
S ome wha t Fa milia r
Not too Fa milia r
Not a t a ll Fa milia r
3.7 P(E) = 1 – (.2 + .15 + .3 + .2) = .15
The sum of the probabilities of the individual outcomes sum to 1.
3.11
a.
2,500
(1) PM   10,000  .25
4,000
(2) PV   10,000  .40
1,000
(3) PM V   10,000  .10
b.
c.
M
M
Total
V
1,000
3,000
4,000
V
1,500
4,500
6,000
Total
2,500
7,500
10,000
(1) PM  V   PM   PV   PM  V   .25  .40  .10  .55
4,500
(2) P M V   10,000  .45

 

(3) P M V  P M V  .15  .30  .45