Question 1:
a. Let
X
be the number of homes (out of 6) in Riverband area of Edmonton that have a
security system. Then
X
∼
Binomial
(6
,
0
.
4). Therefore
P
(
X
= 3) =
6!
3!3!
(0
.
4)
3
(0
.
6)
3
= 0
.
27648
b. See the table below and we have the mean is 1110 and the standard deviation is
√
24900 = 157
.
80.
Admissions Probability
xp
(
x
) (
x

μ
)
2
p
(
x
)
1000
0.6
600
7260
1200
0.3
360
2430
1500
0.1
150
15210
1110
24900
Question 2:
a.
minimum
= 510 and
maximum
= 970
Position of
Q
1
= (36 + 1)
*
0
.
25 = 9
.
25
⇐
Q
1
=
between
Position of ˜
x
= (36 + 1)
*
0
.
5 = 18
.
5
⇐
˜
x
=
between
Position of
Q
3
= (36 + 1)
*
0
.
75 = 27
.
75
⇐
Q
3
=
between
IQR
= 875

735 = 140
Inner fence = (
Q
1

1
.
5
IQR, Q
3
+ 1
.
5
IQR
) = (525
,
1085)
Since all data points are within the inner fence, then there’s no outliers and the boxplot
is just the simple boxplot formed from the 5number summary (510
,
735
,
820
,
875
,
970).
b. No. Empirical Rule should applied to only “bellshaped” distributed data. From the
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