i. The mean is
E [ X ] = ∑ x × P ( X = x ) = ( 44 × 0.04 ) + ( 45 × 0.13 ) + ( 46 × 0.21 ) + ( 47 × 0.29 ) + ( 48 × 0.20 ) + ( 49 × 0.10 ) + ( 50 × 0.03 ) = 1.76 + 5.86 + 9.66 + 13.63 + 9.60 + 4.90 + 1.5 = 46.91 E[X] = \sum x \times P(X=x) \\
= (44 \times 0.04) + (45 \times 0.13) + (46 \times 0.21) + (47 \times 0.29) + (48 \times 0.20) + (49 \times 0.10) + (50 \times 0.03) \\
= 1.76 + 5.86 + 9.66 + 13.63 + 9.60 + 4.90 + 1.5 \\
= 46.91 E [ X ] = ∑ x × P ( X = x ) = ( 44 × 0.04 ) + ( 45 × 0.13 ) + ( 46 × 0.21 ) + ( 47 × 0.29 ) + ( 48 × 0.20 ) + ( 49 × 0.10 ) + ( 50 × 0.03 ) = 1.76 + 5.86 + 9.66 + 13.63 + 9.60 + 4.90 + 1.5 = 46.91
ii.
E [ X 2 ] = ∑ x 2 × P ( X = x ) = 4 4 2 × 0.04 + 4 5 2 × 0.13 + 4 6 2 × 0.21 + 4 7 2 × 0.29 + 4 8 2 × 20 + 4 9 2 × 0.10 + 5 0 2 × 0.03 = 2201.56 V a r [ X ] = E [ X 2 ] – ( E [ X ] ) 2 = 2201.56 − 46. 9 2 = 1.95 E[X^2] = \sum x^2 \times P(X=x)
= 44^2 \times 0.04 + 45^2 \times 0.13 + 46^2 \times 0.21 + 47^2 \times 0.29 + 48^2 \times 20 + 49^2 \times 0.10 + 50^2 \times 0.03 \\
= 2201.56 \\
Var[X] = E[X^2] – (E[X])^2 \\
= 2201.56 -46.9^2 \\
= 1.95 E [ X 2 ] = ∑ x 2 × P ( X = x ) = 4 4 2 × 0.04 + 4 5 2 × 0.13 + 4 6 2 × 0.21 + 4 7 2 × 0.29 + 4 8 2 × 20 + 4 9 2 × 0.10 + 5 0 2 × 0.03 = 2201.56 Va r [ X ] = E [ X 2 ] – ( E [ X ] ) 2 = 2201.56 − 46. 9 2 = 1.95
Standard deviation is
S D [ X ] = V a r [ X ] = 1.3964 SD[X] = \sqrt{Var[X]} = 1.3964 S D [ X ] = Va r [ X ] = 1.3964
iii. Proper function C = 75 + 2X
a) The mean cost:
E [ C ] = 75 + 2 × 46.91 = 168.82 E[C] = 75 + 2 \times 46.91 = 168.82 E [ C ] = 75 + 2 × 46.91 = 168.82
b) Standard deviation is
S D [ C ] = V a r [ C ] = V a r ( 75 + 2 X ) = 2 2 × V a r [ X ] = 2 × S D [ X ] = 2 × 1.3964 = 2.793 SD[C] = \sqrt{Var[C]} \\
= \sqrt{Var(75 + 2X)} \\
= \sqrt{2^2 \times Var[X]} \\
= 2 \times SD[X] \\
= 2 \times 1.3964 \\
= 2.793 S D [ C ] = Va r [ C ] = Va r ( 75 + 2 X ) = 2 2 × Va r [ X ] = 2 × S D [ X ] = 2 × 1.3964 = 2.793
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