Stat 211  Tutorial 3 Solution/Answers
1.
(a)
x
5
5
20
P
(
X
=
x
)
25/36
10/36
1/36
(b)
E
(
X
) =

$1
.
53
and
V ar
(
X
) = 33
.
08256
.
2.
(a)
x
100
200
P
(
X
=
x
)
0.6
0.4
(b) mean,
μ
x
= $140
; variance,
σ
2
x
= 2400
; standard deviation,
σ
x
= $48
.
99
.
(c) mean,
μ
y
= $147
; variance,
σ
2
y
= 2646
; standard deviation,
σ
y
= $51
.
44
.
(d) mean,
μ
p
= $150
; variance,
σ
2
p
= 2400
; standard deviation,
σ
p
= $48
.
99
.
(e) mean,
μ
m
= $140
; variance,
σ
2
m
= 1200
; standard deviation,
σ
m
= $34
.
64
.
(f)
i. $144
ii.
E
(
X
) = 200

100
p
iii.
ˆ
p
= 0
.
56
.
3. 95% of parts produced by a machine in a day are within the required specification.
In a given day, ten parts are randomly selected. Find the probability that
(a) exactly 1 part is not within specification.
Solution
:
X
= number of parts in a set of 10 to be within specification.
X
∼
Binomial
(10
,
0
.
95)
.
If exactly 1 part is not within specification, then exactly 9 parts must be within
specification. Therefore, the required probability is
P
(
X
= 9) = 0
.
3151
(b) at least one part is within specification.
Solution
:
P
(
X
≥
1) = 1

P
(
X <
1) = 1

P
(
X
= 0)
≈
1
.
(c) Let
X
denote the number of parts in the sample of 10 that are within specifica
tion. Find
E
(
X
)
,
E
(2
X
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 Winter '12
 Javid
 Standard Deviation, Variance, X Games, Federer

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