Brienna Matson Assignment 9
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Brienna Matson Seth Eikrem STAT243 Assignment 9
5-2 Pg.245 #28 Construct a probability distribution for a family of 3 children. Let X=the number of boys.
No Boys
One Boy
GGG
GGB
BBG
1/8
1/8
1/8
Total: 1/8
Two Boys GBG
BBG
1/8
1/8
Total: 3/8
Three Boys
BGB
GBB
BBB
1/8
1/8
1/8
Total: 3/8
Total: 1/8
Number of Boys X
0
1
2
3
Probability P(X)
1/8
3/8
3/8
1/8
5-3 Pg.253 # 8 Number of Televisions X
1
2
3
4
Probability P(X)
0.32
0.51
0.12
0.05
MEAN ≈ 2.3 Televisions (X₁×0.32)+(X₂×0.51) + (X₃×0.12) + (X₄×0.05) = (1×0.32) + (2×0.51) + (3×0.12) + (4×0.05) =0.32+1.02+0.36+0.56 =2.26 ≈2.3 Televisions
VARIANCE ≈ -1.1 (1²×0.32)+ (2² × 0.51) + (3² × 0.12) + (4² × 0.05) = 4.24 4.24 – 2.3² = -1.05 ≈ -1.1
STANDARD DEVIATION ≈ 1 1.1 =1.04880884
≈1 I would send 3 program journals to each household because most people have between 2 and 3 televisions in their home.
5-3 Pg.254 # 18 $100,000 - $360 = $99,640 1 - 0.999057 = 0.000943
LADY DIES LADY LIVES
Gain X
$99,640
-$360
Probability P(X)
.000943
.999057
The expected value of the policy for the insurance company is $265.70.
E(X) = -$265.70 (99,640×0.000943) + (-360 × 0.999057) = -265.7 = -$265.70
5-2 Pg.245 #28 Construct a probability distribution for a family of 3 children. Let X=the number of boys.
No Boys
One Boy
GGG
GGB
BBG
1/8
1/8
1/8
Total: 1/8
Two Boys GBG
BBG
1/8
1/8
Total: 3/8
Three Boys
BGB
GBB
BBB
1/8
1/8
1/8
Total: 3/8
Total: 1/8
Number of Boys X
0
1
2
3
Probability P(X)
1/8
3/8
3/8
1/8
5-3 Pg.253 # 8 Number of Televisions X
1
2
3
4
Probability P(X)
0.32
0.51
0.12
0.05
MEAN ≈ 2.3 Televisions (X₁×0.32)+(X₂×0.51) + (X₃×0.12) + (X₄×0.05) = (1×0.32) + (2×0.51) + (3×0.12) + (4×0.05) =0.32+1.02+0.36+0.56 =2.26 ≈2.3 Televisions
VARIANCE ≈ -1.1 (1²×0.32)+ (2² × 0.51) + (3² × 0.12) + (4² × 0.05) = 4.24 4.24 – 2.3² = -1.05 ≈ -1.1
STANDARD DEVIATION ≈ 1 1.1 =1.04880884
≈1 I would send 3 program journals to each household because most people have between 2 and 3 televisions in their home.
5-3 Pg.254 # 18 $100,000 - $360 = $99,640 1 - 0.999057 = 0.000943
LADY DIES LADY LIVES
Gain X
$99,640
-$360
Probability P(X)
.000943
.999057
The expected value of the policy for the insurance company is $265.70.
E(X) = -$265.70 (99,640×0.000943) + (-360 × 0.999057) = -265.7 = -$265.70
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