Random Variables

2.1 Discrete Random Variables

2.1.1 (a) Since

0.08 + 0.11 + 0.27 + 0.33 + P (X = 4) = 1 it follows that

P (X = 4) = 0.21.

(c) F (0) = 0.08 F (1) = 0.19 F (2) = 0.46 F (3) = 0.79 F (4) = 1.00

2.1.2

x_i -4 -1 0 2 3 7

pi 0.21 0.11 0.07 0.29 0.13 0.19

2.1.3

x_i 1 2 3 4 5 6 8 9 10

p_{i 36}¹ ₃₆² ₃₆² ₃₆³ ₃₆² ₃₆⁴ ₃₆² ₃₆¹ ₃₆²

F (xi) ₃₆¹ ₃₆³ ₃₆⁵ ₃₆^{8 10}₃₆ ¹⁴₃₆ ¹⁶₃₆ ¹⁷₃₆ ¹⁹₃₆

49

xi 12 15 16 18 20 24 25 30 36

p_{i 36}⁴ ₃₆² ₃₆¹ ₃₆² ₃₆² ₃₆² ₃₆¹ ₃₆² ₃₆¹

F (xi) ²³₃₆ ²⁵₃₆ ²⁶₃₆ ²⁸₃₆ ³⁰₃₆ ³²₃₆ ³³₃₆ ³⁵₃₆ 1

2.1.4 (a)

x_i 0 1 2

pi 0.5625 0.3750 0.0625

(b)

x_i 0 1 2

F (xi) 0.5625 0.9375 1.000

(c) The value x = 0 is the most likely.

Without replacement:

x_i 0 1 2

pi 0.5588 0.3824 0.0588

F (x_i) 0.5588 0.9412 1.000

Again, x = 0 is the most likely value.

2.1. DISCRETE RANDOM VARIABLES 51

2.1.5

xi -5 -4 -3 -2 -1 0 1 2 3 4 6 8 10 12

p_{i 36}¹ ₃₆¹ ₃₆² ₃₆² ₃₆³ ₃₆³ ₃₆² ₃₆⁵ ₃₆¹ ₃₆⁴ ₃₆³ ₃₆³ ₃₆³ ₃₆³

F (xi) ₃₆¹ ₃₆² ₃₆⁴ ₃₆⁶ ₃₆^{9 12}₃₆ ¹⁴₃₆ ¹⁹₃₆ ²⁰₃₆ ²⁴₃₆ ²⁷₃₆ ³⁰₃₆ ³³₃₆ 1

2.1.6 (a)

xi -6 -4 -2 0 2 4 6

p_i ¹₈ ¹₈ ¹₈ ²₈ ¹₈ ¹₈ ¹₈

(b)

x_i -6 -4 -2 0 2 4 6

F (xi) ¹₈ ²₈ ³₈ ⁵₈ ⁶₈ ⁷₈ 1

(c) The most likely value is x = 0.

2.1.7 (a)

xi 0 1 2 3 4 6 8 12

p_i 0.061 0.013 0.195 0.067 0.298 0.124 0.102 0.140

(b)

xi 0 1 2 3 4 6 8 12

F (x_i) 0.061 0.074 0.269 0.336 0.634 0.758 0.860 1.000

(c) The most likely value is 4.

P(not shipped) = P(X ≤ 1) = 0.074

is a possible set of probability values.

However, since P∞

i=11 i

does not converge, it follows that P (X = i) = ^c_i

is not a possible set of probability values.

2.1.11 (a) The state space is {3, 4, 5, 6}.

(b) P (X = 3) = P (M M M ) = ³₆ ײ₅ ×¹₄ = ₂₀¹

P (X = 4) = P (M M T M ) + P (M T M M ) + P (T M M M ) = ₂₀³ P (X = 5) = P (M M T T M ) + P (M T M T M ) + P (T M M T M )

2.1. DISCRETE RANDOM VARIABLES 53

+ P (M T T M M ) + P (T M T M M ) + P (T T M M M ) = ₂₀⁶ Finally,

P (X = 6) = ¹₂

since the probabilities sum to one, or since the final appointment made is equally likely to be on a Monday or on a Tuesday.

P (X ≤ 3) = ₂₀¹ P (X ≤ 4) = ₂₀⁴ P (X ≤ 5) = ¹⁰₂₀ P (X ≤ 6) = 1

2.2 Continuous Random Variables

2.2.1 (a) Continuous (b) Discrete

(c) Continuous (d) Continuous

(e) Discrete

(f) This depends on what level of accuracy to which it is measured.

It could be considered to be either discrete or continuous.

2.2.2 (b) ^R₄⁶_{x ln(1.5)}¹ dx = _ln(1.5)¹ × [ln(x)]⁶₄

= _ln(1.5)¹ × (ln(6) − ln(4)) = 1.0 (c) P (4.5 ≤ X ≤ 5.5) =^R_4.5^5.5 _{x ln(1.5)}¹ dx

= _ln(1.5)¹ × [ln(x)]^5.5_4.5

= _ln(1.5)¹ × (ln(5.5) − ln(4.5)) = 0.495 (d) F (x) =^R₄^x _{y ln(1.5)}¹ dy

= _ln(1.5)¹ × [ln(y)]^x₄

= _ln(1.5)¹ × (ln(x) − ln(4)) for 4 ≤ x ≤ 6

2.2.3 (a) Since R0

−2

15

64+₆₄^xdx = ₁₆⁷ and

R3 0

3

8 + cx^dx = ⁹₈ +^9c₂ it follows that

7

16 +⁹₈+ ^9c₂ = 1 which gives c = −¹₈.

(b) P (−1 ≤ X ≤ 1) =^R₋₁^{0 15}₆₄+₆₄^xdx +^R₀^13₈ −^x₈^dx

= ₁₂₈⁶⁹

2.2. CONTINUOUS RANDOM VARIABLES 55

(c) F (x) =^R₋₂^{x 15}₆₄+₆₄^{y }dy

= ₁₂₈^x2 +^15x₆₄ +₁₆⁷ for −2 ≤ x ≤ 0

F (x) = ₁₆⁷ +^R₀^x3₈ −^y₈^dy

= −^x₁₆² +^3x₈ +₁₆⁷ for 0 ≤ x ≤ 3

2.2.4 (b) P (X ≤ 2) = F (2) = ¹₄

(c) P (1 ≤ X ≤ 3) = F (3) − F (1)

= ₁₆⁹ −₁₆¹ = ¹₂ (d) f (x) = ^{dF (x)}_dx = ^x₈

for 0 ≤ x ≤ 4

2.2.5 (a) Since F (∞) = 1 it follows that A = 1.

Then F (0) = 0 gives 1 + B = 0 so that B = −1 and F (x) = 1 − e^−x.

(b) P (2 ≤ X ≤ 3) = F (3) − F (2)

= e⁻²− e⁻³ = 0.0855 (c) f (x) = ^{dF (x)}_dx = e^−x

for x ≥ 0

2.2.6 (a) Since R0.5

0.125 A (0.5 − (x − 0.25)²) dx = 1 it follows that A = 5.5054.

(b) F (x) =^R_0.125^x f (y) dy

= 5.5054^x₂ − ^(x−0.25)₃ ³ − 0.06315^ for 0.125 ≤ x ≤ 0.5

(c) F (0.2) = 0.203

2.2.7 (a) Since

F (0) = A + B ln(2) = 0 and

F (10) = A + B ln(32) = 1

it follows that A = −0.25 and B = _ln(16)¹ = 0.361.

(b) P (X > 2) = 1 − F (2) = 0.5 (c) f (x) = ^{dF (x)}_dx = _3x+2^1.08

for 0 ≤ x ≤ 10

2.2.8 (a) Since R10

0 A (e^10−θ− 1) dθ = 1 it follows that

A = (e¹⁰− 11)⁻¹= 4.54 × 10⁻⁵. (b) F (θ) =^R₀^θf (y) dy

= ^e10−θ−e_e10−11^10−θ

for 0 ≤ θ ≤ 10 (c) 1 − F (8) = 0.0002

2.2.9 (a) Since F (0) = 0 and F (50) = 1

it follows that A = 1.0007 and B = −125.09.

(b) P (X ≤ 10) = F (10) = 0.964

(c) P (X ≥ 30) = 1 − F (30) = 1 − 0.998 = 0.002 (d) f (r) = ^{dF (r)}_dr = _(r+5)^375.34

for 0 ≤ r ≤ 50

2.2.10 (a) F (200) = 0.1

(b) F (700) − F (400) = 0.65

2.2. CONTINUOUS RANDOM VARIABLES 57

2.2.11 (a) Since R11

10 Ax(130 − x²) dx = 1 it follows that

A =₈₁₉⁴ .

(b) F (x) =^R₁₀^{x 4y(130−y}₈₁₉ ²⁾ dy

= ₈₁₉^{4 }65x²−^x₄⁴ − 4000^ for 10 ≤ x ≤ 11

(c) F (10.5) − F (10.25) = 0.623 − 0.340 = 0.283

2.3 The Expectation of a Random Variable

2.3.1 E(X) = (0 × 0.08) + (1 × 0.11) + (2 × 0.27) + (3 × 0.33) + (4 × 0.21)

= 2.48

2.3.2 E(X) =^1 ×₃₆^1+^2 ×₃₆^{2 }+^3 ×₃₆^2+^4 ×₃₆^3+^5 ×₃₆^{2 }+^6 ×₃₆^4 + ^8 ×₃₆^2+^9 ×₃₆^1+^10 ×₃₆^{2 }+^12 ×₃₆^4+^15 ×₃₆^{2 }+^16 ×₃₆^1 + ^18 × ₃₆^2+^20 ×₃₆^{2 }+^24 × ₃₆^2+^25 ×₃₆^{1 }+^30 × ₃₆^2+^36 ×₃₆^{1 }

= 12.25

2.3.3 With replacement:

E(X) = (0 × 0.5625) + (1 × 0.3750) + (2 × 0.0625)

= 0.5

Without replacement:

E(X) = (0 × 0.5588) + (1 × 0.3824) + (2 × 0.0588)

= 0.5

2.3.4 E(X) =^1 ײ₅^+^2 ×₁₀^3+^3 ×¹₅^+^4 ×₁₀^{1 }

= 2

2.3.5

xi 2 3 4 5 6 7 8 9 10 15

p_{i 13}¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃¹ ₁₃⁴

E(X) =^2 ×₁₃^1+^3 ×₁₃^{1 }+^4 ×₁₃^1+^5 ×₁₃^1+^6 ×₁₃^{1 } + ^7 ×₁₃^1+^8 ×₁₃^1+^9 ×₁₃^{1 }+^10 × ₁₃^1+^15 ×₁₃^{4 }

= $8.77

If $9 is paid to play the game, the expected loss would be 23 cents.

2.3. THE EXPECTATION OF A RANDOM VARIABLE 59

2.3.6

xi 1 2 3 4 5 6 7 8 9 10 11 12

p_{i 72}⁶ ₇₂⁷ ₇₂⁸ ₇₂^{9 10}₇₂ ¹¹_{72 72}⁶ ₇₂⁵ ₇₂⁴ ₇₂³ ₇₂² ₇₂¹

E(X) =^1 ×₇₂^6+^2 ×₇₂^{6 }+^3 ×₇₂^6+^4 ×₇₂^6+^5 ×₇₂^{6 }+^6 ×₇₂^6 + ^7 ×₇₂^6+^8 ×₇₂^6+^9 ×₇₂^{6 }+^10 × ₇₂^6+^11 ×₇₂^{6 }+^12 × ₇₂^6

= 5.25

2.3.7 P (three sixes are rolled) = ¹₆ ×¹₆× ¹₆

= ₂₁₆¹ so that

E(net winnings) =^−$1 × ²¹⁵₂₁₆^+^$499 ×₂₁₆^{1 }

= $1.31.

If you can play the game a large number of times then you should play the game as often as you can.

2.3.8 The expected net winnings will be negative.

2.3.9

xi 0 1 2 3 4 5

p_i 0.1680 0.2816 0.2304 0.1664 0.1024 0.0512

E(payment) = (0 × 0.1680) + (1 × 0.2816) + (2 × 0.2304) + (3 × 0.1664) + (4 × 0.1024) + (5 × 0.0512)

= 1.9072

E(winnings) = $2 − $1.91 = $0.09

The expected winnings increase to 9 cents per game.

Increasing the probability of scoring a three reduces the expected value of the difference in the scores of the two dice.

2.3.10 (a) E(X) =^R₄⁶x _{x ln(1.5)}¹ dx = 4.94 (b) Solving F (x) = 0.5 gives x = 4.90.

2.3.11 (a) E(X) =^R₀⁴ x ^x₈ dx = 2.67 (b) Solving F (x) = 0.5 gives x =√

8 = 2.83.

2.3.12 E(X) =^R_0.125^0.5 x 5.5054 (0.5 − (x − 0.25)²) dx = 0.3095 Solving F (x) = 0.5 gives x = 0.3081.

2.3.13 E(X) =^R₀¹⁰_e10^θ−11 (e^10−θ − 1) dθ = 0.9977 Solving F (θ) = 0.5 gives θ = 0.6927.

2.3.14 E(X) =^R₀^{50 375.3 r}_(r+5)4 dr = 2.44 Solving F (r) = 0.5 gives r = 1.30.

2.3.15 Let f (x) be a probability density function that is symmetric about the point µ, so that f (µ + x) = f (µ − x).

Then

E(X) =^R_−∞^∞ xf (x) dx

which under the transformation x = µ + y gives E(X) =^R_−∞^∞ (µ + y)f (µ + y) dy

= µ ^R_−∞^∞ f (µ + y) dy + ^R₀^∞ y (f (µ + y) − f (µ − y)) dy

= (µ × 1) + 0 = µ.

2.3.16 E(X) = (3 ×₂₀¹) + (4 × ₂₀³) + (5 × ₂₀⁶) + (6 × ¹⁰₂₀)

= ¹⁰⁵₂₀ = 5.25

2.3.17 (a) E(X) =^R₁₀^{114x2(130−x}₈₁₉ ²⁾ dx

= 10.418234

2.3. THE EXPECTATION OF A RANDOM VARIABLE 61

(b) Solving F (x) = 0.5 gives the median as 10.385.

2.3.18 (a) Since R3

2 A(x − 1.5)dx = 1 it follows that A^x²− 1.5x^3₂= 1 so that A = 1.

(b) Let the median be m.

Then Rm

2 (x − 1.5)dx = 0.5 so that

x²− 1.5x^m₂ = 0.5 which gives

0.5m²− 1.5m + 1 = 0.5.

Therefore, m²− 3m + 1 = 0 so that

m = ^3±

√5 2 .

Since 2 ≤ m ≤ 3 it follows that m = ³⁺

√5

2 = 2.618.

2.4 The Variance of a Random Variable

2.4. THE VARIANCE OF A RANDOM VARIABLE 63

and σ = 1.41.

A small variance is generally preferable if the expected winnings are positive.

2.4.5 (a) E(X²) =^R₄⁶ x²_{x ln(1.5)}¹ dx = 24.66 Then E(X) = 4.94 so that

Var(X) = 24.66 − 4.94²= 0.25.

(b) σ =√

0.25 = 0.5

(c) Solving F (x) = 0.25 gives x = 4.43.

Solving F (x) = 0.75 gives x = 5.42.

(d) The interquartile range is 5.42 − 4.43 = 0.99.

2.4.6 (a) E(X²) =^R₀⁴x^{2 x}₈
dx = 8 Then E(X) = ⁸₃ so that Var(X) = 8 −^8₃^2 = ⁸₉.

(b) σ =^q8₉ = 0.94

(c) Solving F (x) = 0.25 gives x = 2.

Solving F (x) = 0.75 gives x =√

12 = 3.46.

(d) The interquartile range is 3.46 − 2.00 = 1.46.

2.4.7 (a) E(X²) =^R_0.125^0.5 x² 5.5054 (0.5 − (x − 0.25)²) dx = 0.1073 Then E(X) = 0.3095 so that

Var(X) = 0.1073 − 0.3095² = 0.0115.

(b) σ =√

0.0115 = 0.107

(c) Solving F (x) = 0.25 gives x = 0.217.

Solving F (x) = 0.75 gives x = 0.401.

(d) The interquartile range is 0.401 − 0.217 = 0.184.

2.4.8 (a) E(X²) =^R₀¹⁰_e10^θ−11² (e^10−θ− 1) dθ

= 1.9803

Then E(X) = 0.9977 so that Var(X) = 1.9803 − 0.9977² = 0.985.

(b) σ =√

0.985 = 0.992

(c) Solving F (θ) = 0.25 gives θ = 0.288.

Solving F (θ) = 0.75 gives θ = 1.385.

(d) The interquartile range is 1.385 − 0.288 = 1.097.

2.4.9 (a) E(X²) =^R₀^{50 375.3 r}_(r+5)4² dr = 18.80 Then E(X) = 2.44 so that Var(X) = 18.80 − 2.44²= 12.8.

(b) σ =√

12.8 = 3.58

(c) Solving F (r) = 0.25 gives r = 0.50.

Solving F (r) = 0.75 gives r = 2.93.

(d) The interquartile range is 2.93 − 0.50 = 2.43.

2.4.10 Adding and subtracting two standard deviations from the mean value gives:

P (60.4 ≤ X ≤ 89.6) ≥ 0.75

Adding and subtracting three standard deviations from the mean value gives:

P (53.1 ≤ X ≤ 96.9) ≥ 0.89

2.4.11 The interval (109.55, 112.05) is (µ − 2.5c, µ + 2.5c) so Chebyshev’s inequality gives:

P (109.55 ≤ X ≤ 112.05) ≥ 1 −_2.5¹2 = 0.84

2.4.12 E(X²) =^3²×₂₀^{1 }+^4²× ₂₀^3+^5²×₂₀^6+^6²×¹⁰₂₀^

= ⁵⁶⁷₂₀

Var(X) = E(X²) − (E(X))²

2.4. THE VARIANCE OF A RANDOM VARIABLE 65

= ⁵⁶⁷₂₀ −^105₂₀^2 = ⁶³₈₀

The standard deviation is^p63/80 = 0.887.

2.4.13 (a) E(X²) =^R₁₀^{114x3(130−x}₈₁₉ ²⁾ dx

= 108.61538 Therefore,

Var(X) = E(X²) − (E(X))² = 108.61538 − 10.418234² = 0.0758 and the standard deviation is√

0.0758 = 0.275.

(b) Solving F (x) = 0.8 gives the 80th percentile of the resistance as 10.69, and solving F (x) = 0.1 gives the 10th percentile of the resistance as 10.07.

2.4.14 (a) Since

1 =^R₂³Ax^2.5 dx = _3.5^A × (3^3.5− 2^3.5) it follows that A = 0.0987.

(b) E(X) =^R₂³0.0987 x^3.5 dx

= ^0.0987_4.5 × (3^4.5− 2^4.5) = 2.58 (c) E(X²) =^R₂³0.0987 x^4.5 dx

= ^0.0987_5.5 × (3^5.5− 2^5.5) = 6.741 Therefore,

Var(X) = 6.741 − 2.58²= 0.085 and the standard deviation is√

0.085 = 0.29.

(d) Solving

0.5 =^R₂^x0.0987 y^2.5 dy

= ^0.0987_3.5 × (x^3.5− 2^3.5) gives x = 2.62.

2.4.15 E(X) = (−1 × 0.25) + (1 × 0.4) + (4 × 0.35)

= $1.55

E(X²) = ((−1)²× 0.25) + (1²× 0.4) + (4²× 0.35)

= 6.25

Therefore, the variance is

E(X²) − (E(X))²= 6.25 − 1.55²= 3.8475 and the standard deviation is√

3.8475 = $1.96. and the standard deviation is√

0.0834 = 0.289. and the standard deviation is√

0.7124 = 0.844.

2.4. THE VARIANCE OF A RANDOM VARIABLE 67

2.4.18 (a) E(X) =^R₋₁^{1 x(1−x)}₂ dx = −¹₃ (b) E(X²) =^R₋₁^{1 x2(1−x)}₂ dx = ¹₃

Therefore,

Var(X) = E(X²) − (E(X))² = ¹₃ −¹₉ = ²₉ and the standard deviation is

√ 2

3 = 0.471.

(c) Solving Ry

−1 (1−x)

2 dx = 0.75 gives y = 0.

2.5 Jointly Distributed Random Variables

2.5. JOINTLY DISTRIBUTED RANDOM VARIABLES 69

(b) p_1|X=2 = P (Y = 1|X = 2) = _p^p21

2+ = ^0.08_0.24 = ₂₄⁸ p_2|X=2 = P (Y = 2|X = 2) = _p^p22

2+ = ^0.15_0.24 = ¹⁵₂₄ p_3|X=2 = P (Y = 3|X = 2) = _p^p23

2+ = ^0.01_0.24 = ₂₄¹ E(Y |X = 2) =^1 ×₂₄^8+^2 ×¹⁵₂₄^+^3 ×₂₄^{1 }

= ⁴¹₂₄ = 1.7083

E(Y²|X = 2) =^1²×₂₄^8+^2²×¹⁵₂₄^+^3²× ₂₄^1

= ⁷⁷₂₄ = 3.2083

Var(Y |X = 2) = E(Y²|X = 2) − E(Y |X = 2)²

= 3.2083 − 1.7083²= 0.290 σ_{Y |X=2} =√

0.290 = 0.538

2.5.3 (a) Since R3

x=−2

R6

y=4 A(x − 3)y dx dy = 1 it follows that A = −₁₂₅¹ . (b) P (0 ≤ X ≤ 1, 4 ≤ Y ≤ 5)

=^R_x=0^{1 R}_y=4^{5 (3−x)y}₁₂₅ dx dy

= ₁₀₀⁹

(c) f_X(x) =^R₄^{6 (3−x)y}₁₂₅ dy = ^2(3−x)₂₅ for −2 ≤ x ≤ 3

f_Y(y) =^R₋₂^{3 (3−x)y}₁₂₅ dx = ₁₀^y for 4 ≤ x ≤ 6

(d) The random variables X and Y are independent since fX(x) × fY(y) = f (x, y)

and the ranges of the random variables are not related.

(e) Since the random variables are independent it follows that f_{X|Y =5}(x) is equal to fX(x).

2.5.4 (a)

X\Y 0 1 2 3 pi+

0 ₁₆¹ ₁₆¹ 0 0 ₁₆² 1 ₁₆¹ ₁₆³ ₁₆² 0 ₁₆⁶ 2 0 ₁₆² ₁₆³ ₁₆¹ ₁₆⁶ 3 0 0 ₁₆¹ ₁₆¹ ₁₆²

p+j 2 16

6 16

6 16

2

16 1

(b) See the table above.

(c) The random variables X and Y are not independent.

For example, notice that

p₀₊× p₊₀ = ₁₆² ×₁₆² = ¹₄ 6= p₀₀= ₁₆¹ .

(d) E(X) =^0 ×₁₆^{2 }+^1 ×₁₆^6+^2 ×₁₆^6+^3 ×₁₆^{2 }= ³₂ E(X²) =^0²×₁₆^{2 }+^1²× ₁₆^6+^2²×₁₆^6+^3²×₁₆^2= 3

Var(X) = E(X²) − E(X)²= 3 −^3₂^2= ³₄

The random variable Y has the same mean and variance as X.

(e) E(XY ) =^1 × 1 ×₁₆^{3 }+^1 × 2 ×₁₆^2+^2 × 1 ×₁₆^{2 }

+ ^2 × 2 × ₁₆^3+^2 × 3 ×₁₆^{1 }+^3 × 2 × ₁₆^1+^3 × 3 ×₁₆^{1 }

= ⁴⁴₁₆

Cov(X, Y ) = E(XY ) − (E(X) × E(Y ))

= ⁴⁴₁₆−^3₂ ׳₂^= ¹₂

(f) P (X = 0|Y = 1) = _p^p01

+1 = (₁₆¹) (₁₆⁶) = ¹₆ P (X = 1|Y = 1) = _p^p11

+1 = (16³) (16⁶) = ¹₂ P (X = 2|Y = 1) = _p^p21

+1 = (₁₆²) (16⁶) = ¹₃

2.5. JOINTLY DISTRIBUTED RANDOM VARIABLES 71

P (X = 3|Y = 1) = _p^p31

+1 = ⁰

(₁₆⁶) = 0

E(X|Y = 1) =^0 ×¹₆^+^1 ×¹₂^+^2 ×¹₃^+ (3 × 0) = ⁷₆

E(X²|Y = 1) =^0²×¹₆^+^1²×¹₂^+^2²×¹₃^+ 3²× 0
= ¹¹₆ Var(X|Y = 1) = E(X²|Y = 1) − E(X|Y = 1)²

= ¹¹₆ −^7₆^2 = ¹⁷₃₆

2.5.5 (a) Since R2

x=1

R3

y=0 A(e^x+y + e^2x−y) dx dy = 1 it follows that A = 0.00896.

(b) P (1.5 ≤ X ≤ 2, 1 ≤ Y ≤ 2)

=^R_x=1.5^{2 R}_y=1² 0.00896 (e^x+y + e^2x−y) dx dy

= 0.158

(c) fX(x) =^R₀³ 0.00896 (e^x+y + e^2x−y) dy

= 0.00896 (e^x+3− e^2x−3− e^x+ e^2x) for 1 ≤ x ≤ 2

f_Y(y) =^R₁² 0.00896 (e^x+y + e^2x−y) dx

= 0.00896 (e^2+y+ 0.5e^4−y− e^1+y− 0.5e^2−y) for 0 ≤ y ≤ 3

(d) No, since fX(x) × fY(y) 6= f (x, y).

(e) f_{X|Y =0}(x) = ^{f (x,0)}_f

Y(0) = ^ex_28.28^+e2x

2.5.6 (a)

X\Y 0 1 2 pi+

0 ₁₀₂²⁵ ₁₀₂²⁶ ₁₀₂⁶ ₁₀₂⁵⁷ 1 ₁₀₂²⁶ ₁₀₂¹³ 0 ₁₀₂³⁹ 2 ₁₀₂⁶ 0 0 ₁₀₂⁶

p_{+j 102}⁵⁷ ₁₀₂³⁹ ₁₀₂⁶ 1

(b) See the table above.

(c) No, the random variables X and Y are not independent.

For example, p226= p₂₊× p₊₂.

(d) E(X) =^0 ×₁₀₂^57+^1 ×₁₀₂^39+^2 ×₁₀₂^{6 }= ¹₂ E(X²) =^0²×₁₀₂^57+^1²×₁₀₂^39+^2²× ₁₀₂^{6 }= ²¹₃₄

Var(X) = E(X²) − E(X)²= ²¹₃₄ −^1₂^2= ²⁵₆₈

The random variable Y has the same mean and variance as X.

(e) E(XY ) = 1 × 1 × p₁₁= ₁₀₂¹³

Cov(X, Y ) = E(XY ) − (E(X) × E(Y ))

= ₁₀₂¹³ −^1₂ ×¹₂^= −₂₀₄²⁵

(f) Corr(X, Y ) = √ Cov(X,Y )

Var(X)Var(Y ) = −¹₃ (g) P (Y = 0|X = 0) = _p^p00

0+ = ²⁵₅₇ P (Y = 1|X = 0) = _p^p01

0+ = ²⁶₅₇ P (Y = 2|X = 0) = _p^p02

0+ = ₅₇⁶ P (Y = 0|X = 1) = _p^p10

1+ = ²₃ P (Y = 1|X = 1) = _p^p11

1+ = ¹₃

2.5. JOINTLY DISTRIBUTED RANDOM VARIABLES 73

P (Y = 2|X = 1) = _p^p12

1+ = 0 2.5.7 (a)

X\Y 0 1 2 p_i+

0 ₁₆⁴ ₁₆⁴ ₁₆¹ ₁₆⁹ 1 ₁₆⁴ ₁₆² 0 ₁₆⁶ 2 ₁₆¹ 0 0 ₁₆¹

p_{+j 16}⁹ ₁₆⁶ ₁₆¹ 1

(b) See the table above.

(c) No, the random variables X and Y are not independent.

For example, p₂₂6= p₂₊× p₊₂.

(d) E(X) =^0 ×₁₆^{9 }+^1 ×₁₆^6+^2 ×₁₆^1= ¹₂ E(X²) =^0²×₁₆^{9 }+^1²× ₁₆^6+^2²×₁₆^1= ⁵₈ Var(X) = E(X²) − E(X)²= ⁵₈ −^1₂^2 = ³₈ = 0.3676

The random variable Y has the same mean and variance as X.

(e) E(XY ) = 1 × 1 × p₁₁= ¹₈

Cov(X, Y ) = E(XY ) − (E(X) × E(Y ))

= ¹₈−^1₂×¹₂^= −¹₈

(f) Corr(X, Y ) = √ Cov(X,Y )

Var(X)Var(Y ) = −¹₃ (g) P (Y = 0|X = 0) = _p^p00

0+ = ⁴₉ P (Y = 1|X = 0) = _p^p01

0+ = ⁴₉ P (Y = 2|X = 0) = _p^p02

0+ = ¹₉ P (Y = 0|X = 1) = _p^p10

1+ = ²₃

P (Y = 1|X = 1) = _p^p11

1+ = ¹₃ P (Y = 2|X = 1) = _p^p12

1+ = 0 2.5.8 (a) Since

R5 x=0

R5

y=0 A (20 − x − 2y) dx dy = 1 it follows that A = 0.0032

(b) P (1 ≤ X ≤ 2, 2 ≤ Y ≤ 3)

=^R_x=1^{2 R}_y=2³ 0.0032 (20 − x − 2y) dx dy

= 0.0432

(c) f_X(x) =^R_y=0⁵ 0.0032 (20 − x − 2y) dy = 0.016 (15 − x) for 0 ≤ x ≤ 5

f_Y(y) =^R_x=0⁵ 0.0032 (20 − x − 2y) dx = 0.008 (35 − 4y) for 0 ≤ y ≤ 5

(d) No, the random variables X and Y are not independent since f (x, y) 6= f_X(x)f_Y(y).

(e) E(X) =^R₀⁵ x 0.016 (15 − x) dx = ⁷₃ E(X²) =^R₀⁵ x²0.016 (15 − x) dx = ¹⁵₂ Var(X) = E(X²) − E(X)²= ¹⁵₂ −^7₃^2= ³⁷₁₈

(f) E(Y ) =^R₀⁵ y 0.008 (35 − 4y) dy = ¹³₆ E(Y²) =^R₀⁵ y² 0.008 (35 − 4y) dy = ²⁰₃ Var(Y ) = E(Y²) − E(Y )²= ²⁰₃ −^13₆^2 = ⁷¹₃₆

(g) f_{Y |X=3}(y) = ^{f (3,y)}_f

X(3) = ^17−2y₆₀ for 0 ≤ y ≤ 5

(h) E(XY ) =^R_x=0^{5 R}_y=0⁵ 0.0032 xy (20 − x − 2y) dx dy = 5 Cov(X, Y ) = E(XY ) − (E(X) × (EY ))

= 5 −^7₃ ×¹³₆^= −₁₈¹

2.5. JOINTLY DISTRIBUTED RANDOM VARIABLES 75

(i) Corr(X, Y ) = √ Cov(X,Y )

Var(X)Var(Y ) = −0.0276

2.5.9 (a) P (same score) = P (X = 1, Y = 1) + P (X = 2, Y = 2) + P (X = 3, Y = 3) + P (X = 4, Y = 4)

= 0.80

(b) P (X < Y ) = P (X = 1, Y = 2) + P (X = 1, Y = 3) + P (X = 1, Y = 4) + P (X = 2, Y = 3) + P (X = 2, Y = 4) + P (X = 3, Y = 4)

= 0.07 (c)

xi 1 2 3 4

p_i+ 0.12 0.20 0.30 0.38

E(X) = (1 × 0.12) + (2 × 0.20) + (3 × 0.30) + (4 × 0.38) = 2.94 E(X²) = (1²× 0.12) + (2²× 0.20) + (3²× 0.30) + (4²× 0.38) = 9.70 Var(X) = E(X²) − E(X)²= 9.70 − (2.94)²= 1.0564

(d)

yj 1 2 3 4

p_+j 0.14 0.21 0.30 0.35

E(Y ) = (1 × 0.14) + (2 × 0.21) + (3 × 0.30) + (4 × 0.35) = 2.86 E(Y²) = (1²× 0.14) + (2²× 0.21) + (3²× 0.30) + (4²× 0.35) = 9.28 Var(Y ) = E(Y²) − E(Y )²= 9.28 − (2.86)² = 1.1004

(e) The scores are not independent.

For example, p116= p₁₊× p₊₁.

The scores would not be expected to be independent since they apply to the two inspectors’ assessments of the same building. If they were independent it would suggest that one of the inspectors is randomly assigning a safety score without paying any attention to the actual state of the building.

(f) P (Y = 1|X = 3) = _p^p31

3+ = ₃₀¹

P (Y = 2|X = 3) = _p^p32

3+ = ₃₀³ P (Y = 3|X = 3) = _p^p33

3+ = ²⁴₃₀ P (Y = 4|X = 3) = _p^p34

3+ = ₃₀²

(g) E(XY ) =^P4_i=1^P4_j=1 i j p_ij = 9.29 Cov(X, Y ) = E(XY ) − (E(X) × E(Y ))

= 9.29 − (2.94 × 2.86) = 0.8816 (h) Corr(X, Y ) = √Cov(X,Y )

VarX VarY = ^√1.0564×1.1004^0.8816 = 0.82

A high positive correlation indicates that the inspectors are consistent.

The closer the correlation is to one the more consistent the inspectors are.

2.5.10 (a) ^R_x=0^{2 R}_y=0^{2 R}_z=0^{2 3xyz}₃₂² dx dy dz = 1 (b) ^R_x=0^{1 R}_y=0.5^{1.5 R}_z=1^{2 3xyz}₃₂² dx dy dz = ₆₄⁷

(c) f_X(x) =^R_x=0^{2 R}_y=0^{2 3xyz}₃₂² dy dz = ^x₂ for 0 ≤ 2 ≤ x

2.6. COMBINATIONS AND FUNCTIONS OF RANDOM VARIABLES 77

2.6 Combinations and Functions of Random variables

2.6.1 (a) E(3X + 7) = 3E(X) + 7 = 13 Var(3X + 7) = 3²Var(X) = 36 (b) E(5X − 9) = 5E(X) − 9 = 1

Var(5X − 9) = 5²Var(X) = 100

(c) E(2X + 6Y ) = 2E(X) + 6E(Y ) = −14 Var(2X + 6Y ) = 2²Var(X) + 6²Var(Y ) = 88 (d) E(4X − 3Y ) = 4E(X) − 3E(Y ) = 17

Var(4X − 3Y ) = 4²Var(X) + 3²Var(Y ) = 82 (e) E(5X − 9Z + 8) = 5E(X) − 9E(Z) + 8 = −54

Var(5X − 9Z + 8) = 5²Var(X) + 9²Var(Z) = 667 (f) E(−3Y − Z − 5) = −3E(Y ) − E(Z) − 5 = −4

Var(−3Y − Z − 5) = (−3)²Var(Y ) + (−1)²Var(Z) = 25 (g) E(X + 2Y + 3Z) = E(X) + 2E(Y ) + 3E(Z) = 20

Var(X + 2Y + 3Z) = Var(X) + 2²Var(Y ) + 3²Var(Z) = 75 (h) E(6X + 2Y − Z + 16) = 6E(X) + 2E(Y ) − E(Z) + 16 = 14

Var(6X + 2Y − Z + 16) = 6²Var(X) + 2²Var(Y ) + (−1)²Var(Z) = 159

2.6.2 E(aX + b) =^R(ax + b) f (x) dx

= a^R x f (x) dx + b^R f (x) dx

= aE(X) + b

Var(aX + b) = E((aX + b − E(aX + b))²)

= E((aX − aE(X))²)

= a²E((X − E(X))²)

= a²Var(X)

2.6.3 E(Y ) = 3E(X₁) = 3µ Var(Y ) = 3²Var(X₁) = 9σ²

E(Z) = E(X₁) + E(X₂) + E(X₃) = 3µ

Var(Z) = Var(X₁) + Var(X₂) + Var(X₃) = 3σ² The random variables Y and Z have the same mean but Z has a smaller variance than Y .

2.6.4 length = A1+ A2+ B

E(length) = E(A₁) + E(A₂) + E(B) = 37 + 37 + 24 = 98

Var(length) = Var(A₁) + Var(A₂) + Var(B) = 0.7²+ 0.7²+ 0.3²= 1.07

2.6.5 Let the random variable X_i be the winnings from the i^th game.

Then

E(X_i) =^10 ×¹₈^+^(−1) ×⁷₈^= ³₈ and

E(X_i²) =^10²×¹₈^+^(−1)²×⁷₈^= ¹⁰⁷₈ so that

Var(Xi) = E(X_i²) − (E(Xi))² = ⁸⁴⁷₆₄.

The total winnings from 50 (independent) games is Y = X₁+ . . . + X₅₀

and

E(Y ) = E(X1) + . . . + E(X50) = 50 × ³₈ = ⁷⁵₄ = $18.75 with

Var(Y ) = Var(X₁) + . . . + Var(X₅₀) = 50 ×⁸⁴⁷₆₄ = 661.72 so that σ_Y =√

661.72 = $25.72.

2.6.6 (a) E(average weight) = 1.12 kg

Var(average weight) =^0.03₂₅² = 3.6 × 10⁻⁵

2.6. COMBINATIONS AND FUNCTIONS OF RANDOM VARIABLES 79

The standard deviation is ^√0.03

25 = 0.0012 kg.

(b) It is required that ^0.03√_n ≤ 0.005 which is satisfied for n ≥ 36.

2.6.7 Let the random variable Xibe equal to 1 if an ace is drawn on the i^thdrawing (which happens with a probability of ₁₃¹) and equal to 0 if an ace is not drawn on the i^th drawing (which happens with a probability of ¹²₁₃).

Then the total number of aces drawn is Y = X₁+ . . . + X₁₀.

Notice that E(X_i) = ₁₃¹ so that regardless of whether the drawing is performed with or without replacement it follows that

E(Y ) = E(X1) + . . . + E(X10) = ¹⁰₁₃. Also, notice that E(X_i²) = ₁₃¹ so that Var(Xi) = ₁₃¹ −^₁₃^{1 2}= ₁₆₉¹².

If the drawings are made with replacement then the random variables X_i are inde-pendent so that

Var(Y ) = Var(X1) + . . . + Var(X10) = ¹²⁰₁₆₉.

However, if the drawings are made without replacement then the random variables Xi are not independent.

2.6.8 F_X(x) = P (X ≤ x) = x² for 0 ≤ x ≤ 1

(a) FY(y) = P (Y ≤ y) = P (X³ ≤ y) = P (X ≤ y^1/3) = FX(y^1/3) = y^2/3 and so

f_Y(y) =²₃y^−1/3 for 0 ≤ y ≤ 1

E(y) =^R₀¹ y fY(y) dy = 0.4 (b) FY(y) = P (Y ≤ y) = P (√

X ≤ y) = P (X ≤ y²) = FX(y²) = y⁴ and so

f_Y(y) = 4y³ for 0 ≤ y ≤ 1

E(y) =^R₀¹ y f_Y(y) dy = 0.8

(c) F_Y(y) = P (Y ≤ y) = P (_1+X¹ ≤ y) = P (X ≥ ¹_y − 1)

= 1 − F_X(¹_y − 1) = ²_y −_y¹₂

and so

2.6. COMBINATIONS AND FUNCTIONS OF RANDOM VARIABLES 81

Therefore,

F_X(x) = ^x2(3L−2x)_L3

for 0 ≤ x ≤ L.

(b) The random variable corresponding to the difference between the lengths of the two pieces of rod is

W = |L − 2X|.

Therefore,

F_W(w) = P^L₂ −^w₂ ≤ X ≤ ^L₂ +^w₂^= F_X^L₂ +^w₂^− F_X^L₂ −^w₂^

= ^w(3L_2L^2−w3 ²⁾

and

fW(w) = ^3(L_2L^2−w3 ²⁾

for 0 ≤ w ≤ L.

(c) E(W ) =^R₀^L w fW(w) dw = ³₈L

2.6.11 (a) The return has an expectation of $100, a standard deviation of $20, and a variance of 400.

(b) The return has an expectation of $100, a standard deviation of $30, and a variance of 900.

(c) The return from fund A has an expectation of $50, a standard deviation of $10, and a variance of 100.

The return from fund B has an expectation of $50, a standard deviation of $15, and a variance of 225.

Therefore, the total return has an expectation of $100 and a variance of 325, so that the standard deviation is $18.03.

(d) The return from fund A has an expectation of $0.1x, a standard deviation of $0.02x,

and a variance of 0.0004x².

The return from fund B has an expectation of $0.1(1000 − x), a standard deviation of $0.03(1000 − x),

and a variance of 0.0009(1000 − x)².

Therefore, the total return has an expectation of $100 and a variance of 0.0004x²+ 0.0009(1000 − x)².

This variance is minimized by taking x = $692, and the minimum value of the variance is 276.9 which corresponds to a standard deviation of $16.64.

This problem illustrates that the variability of the return on an investment can be reduced by diversifying the investment, so that it is spread over several funds.

2.6.12 The expected value of the total resistance is 5 × E(X) = 5 × 10.418234 = 52.09.

The variance of the total resistance is 5 × Var(X) = 5 × 0.0758 = 0.379 so that the standard deviation is√

0.379 = 0.616. so that the standard deviation is^p95/3 = 5.63.

(b) The mean is so that the standard deviation is√

43.68 = 6.61.

2.6.14 1000 = E(Y ) = a + bE(X) = a + (b × 77) 10² = Var(Y ) = b²Var(X) = b²× 9²

Solving these equations gives a = 914.44 and b = 1.11,

2.6. COMBINATIONS AND FUNCTIONS OF RANDOM VARIABLES 83

or a = 1085.56 and b = −1.11.

2.6.15 (a) The mean is µ = 65.90.

The standard deviation is ^√σ₅ = ^0.32√₅ = 0.143.

(b) The mean is 8µ = 8 × 65.90 = 527.2.

The standard deviation is√

8σ =√

8 × 0.32 = 0.905.

2.6.16 (a) E(A) = ^E(X1)+E(X₂ ²⁾ = ^{W +W}₂ = W Var(A) = Var(X1)+Var(X2)

4 = ³²⁺⁴₄ ² = ²⁵₄ The standard deviation is ⁵₂ = 2.5.

(b) Var(B) = δ²Var(X₁) + (1 − δ)²Var(X₂) = 9δ²+ 16(1 − δ)² This is minimized when δ = ¹⁶₂₅ and the minimum value is ¹⁴⁴₂₅ so that the minimum standard deviation is ¹²₅ = 2.4.

2.6.17 When a die is rolled once the expectation is 3.5 and the standard deviation is 1.71 (see Games of Chance in section 2.4).

Therefore, the sum of eighty die rolls has an expectation of 80 × 3.5 = 280 and a standard deviation of√

80 × 1.71 = 15.3.

2.6.18 (a) The expectation is 4 × 33.2 = 132.8 seconds.

The standard deviation is√

4 × 1.4 = 2.8 seconds.

(b) E(A1+ A2+ A3+ A4− B₁− B₂− B₃− B₄)

= E(A1) + E(A2) + E(A3) + E(A4) − E(B1) − E(B2) − E(B3) − E(B4)

= (4 × 33.2) − (4 × 33.0) = 0.8

Var(A₁+ A₂+ A₃+ A₄− B₁− B₂− B₃− B₄)

= Var(A₁) + Var(A₂) + Var(A₃) + Var(A₄) + Var(B1) + Var(B2) + Var(B3) + Var(B4)

= (4 × 1.4²) + (4 × 1.3²) = 14.6 The standard deviation is√

14.6 = 3.82.

(c) E^A₁−^A2+A₃^3+A4

= E(A1) −^E(A₃²⁾ −^E(A₃³⁾−^E(A₃⁴⁾ = 0 Var^A₁−^A2+A₃^3+A4

= Var(A1) +Var(A2)

9 +Var(A3)

9 +Var(A4) 9

= ⁴₃× 1.4²= 2.613

The standard deviation is√

2.613 = 1.62.

2.6.19 Let X be the temperature in Fahrenheit and let Y be the temperature in Centigrade.

E(Y ) = E^5(X−32)₉ ^=^{5(E(X)−32)}₉ ^=^5(110−32)₉ ^= 43.33

The standard deviation is√

1.49 = 1.22.

2.6.20 Var(0.5X_α+ 0.3X_β+ 0.2X_γ)

= 0.5²Var(Xα) + 0.3²Var(X_β) + 0.2²Var(Xγ)

= (0.5²× 1.2²) + (0.3²× 2.4²) + (0.2²× 3.1²) = 1.26 The standard deviation is√

1.26 = 1.12.

2.6.21 The inequality ^√56_n ≤ 10 is satisfied for n ≥ 32.

2.6.22 (a) E(X₁+ X₂) = E(X₁) + E(X₂) = 7.74

Var(X1+ X2) = Var(X1) + Var(X2) = 0.0648 The standard deviation is√

0.0648 = 0.255.

(b) E(X₁+ X₂+ X₃) = E(X₁) + E(X₂) + E(X₃) = 11.61

Var(X1+ X2+ X3) = Var(X1) + Var(X2) + Var(X3) = 0.0972 The standard deviation is√

0.0972 = 0.312.

The standard deviation is√

0.0081 = 0.09.

2.6. COMBINATIONS AND FUNCTIONS OF RANDOM VARIABLES 85

(d) E^X₃−^X1+X₂ ^2= E(X₃) −^E(X1)+E(X₂ ²⁾ = 0 Var^X3−^X1+X₂ ^2= Var(X3) + Var(X1)+Var(X2)

4 = 0.0486

The standard deviation is√

0.0486 = 0.220.

2.8 Supplementary Problems

2.8. SUPPLEMENTARY PROBLEMS 87

Var(X) = E(X²) − E(X)²= ³⁸³₃₀ −^7₂^2 = ³¹₆₀ (c)

x_i 2 3 4 5

pi 2 10

3 10

3 10

2 10

E(X) =^2 ×₁₀^2+^3 ×₁₀^{3 }+^4 ×₁₀^3+^5 ×₁₀^2= ⁷₂ E(X²) =^2²×₁₀^{2 }+^3²× ₁₀^3+^4²×₁₀^3+^5²×₁₀^2= ¹³³₁₀ Var(X) = E(X²) − E(X)²= ¹³³₁₀ −^7₂^2 = ²¹₂₀

2.8.4 Let X_i be the value of the i^th card dealt.

Then

E(X_i) =^2 ×₁₃^1+^3 ×₁₃^{1 }+^4 ×₁₃^1+^5 ×₁₃^1+^6 ×₁₃^1 + ^7 ×₁₃^1+^8 ×₁₃^1+^9 ×₁₃^{1 }+^10 × ₁₃^1+^15 ×₁₃^{4 }= ¹¹⁴₁₃ The total score of the hand is

Y = X₁+ . . . + X₁₃ which has an expectation

E(Y ) = E(X1) + . . . + E(X13) = 13 × ¹¹⁴₁₃ = 114.

2.8.5 (a) Since R11

1 A^3₂^xdx = 1

it follows that A = _1.5^ln(1.5)11−1.5 = _209.6¹ .

(b) F (x) =^R₁^x _209.6^{1 3}₂^ydy

= 0.01177^3₂^x− 0.01765 for 1 ≤ x ≤ 11

(c) Solving F (x) = 0.5 gives x = 9.332.

(d) Solving F (x) = 0.25 gives x = 7.706.

Solving F (x) = 0.75 gives x = 10.305.

The interquartile range is 10.305 − 7.706 = 2.599.

2.8.6 (a) fX(x) =^R₁²4x(2 − y) dy = 2x for 0 ≤ x ≤ 1 (b) f_Y(y) =^R₀¹4x(2 − y) dx = 2(2 − y) for 1 ≤ y ≤ 2

Since f (x, y) = f_X(x) × f_Y(y) the random variables are independent.

(c) Cov(X, Y ) = 0 because the random variables are independent.

(d) f_{X|Y =1.5}(x) = fX(x) because the random variables are independent.

2.8.7 (a) Since R10

5 A^x +_x^2dx = 1 it follows that A = 0.02572.

(b) F (x) =^R₅^x 0.02572^y +²_y^dy

= 0.0129x²+ 0.0514 ln(x) − 0.404 for 5 ≤ x ≤ 10

(c) E(X) =^R₅¹⁰ 0.02572 x ^x +_x^2dx = 7.759

(d) E(X²) =^R₅¹⁰ 0.02572 x^{2 }x +²_x^dx = 62.21 Var(X) = E(X²) − E(X)²= 62.21 − 7.759² = 2.01 (e) Solving F (x) = 0.5 gives x = 7.88.

(f) Solving F (x) = 0.25 gives x = 6.58.

Solving F (x) = 0.75 gives x = 9.00.

The interquartile range is 9.00 − 6.58 = 2.42.

(g) The expectation is E(X) = 7.759.

The variance is Var(X)

10 = 0.0201.

2.8.8 Var(a₁X₁+ a₂X₂+ . . . + a_nX_n+ b)

= Var(a₁X₁) + . . . + Var(a_nX_n) + Var(b)

= a²₁Var(X1) + . . . + a²_nVar(Xn) + 0

2.8. SUPPLEMENTARY PROBLEMS 89

2.8.9 Y = ⁵₃X − 25

2.8.10 Notice that E(Y ) = aE(X) + b and Var(Y ) = a²Var(X).

Also,

Cov(X, Y ) = E( (X − E(X)) (Y − E(Y )) )

= E( (X − E(X)) a(X − E(X)) )

= aVar(X).

Therefore,

Corr(X, Y ) = √ Cov(X,Y )

Var(X)Var(Y ) = √ ^aVar(X)

Var(X)a²Var(X) = _|a|^a which is 1 if a > 0 and is −1 if a < 0.

2.8.11 The expected amount of a claim is E(X) =^R₀¹⁸⁰⁰ x 972,000,000^x(1800−x) dx = $900.

Consequently, the expected profit from each customer is

$100 − $5 − (0.1 × $900) = $5.

The expected profit from 10,000 customers is therefore 10, 000 × $5 = $50, 000.

The profits may or may not be independent depending on the type of insurance and the pool of customers.

If large natural disasters affect the pool of customers all at once then the claims would not be independent.

2.8.12 (a) The expectation is 5 × 320 = 1600 seconds.

The variance is 5 × 63² = 19845 and the standard deviation is√

19845 = 140.9 seconds.

(b) The expectation is 320 seconds.

The variance is ⁶³₁₀² = 396.9 and the standard deviation is√

396.9 = 19.92 seconds.

2.8.13 (a) The state space is the positive integers from 1 to n, with each outcome having a probability value of _n¹. (b) E(X) =^_n¹ × 1^+^_n¹ × 2^+ . . . +^1_n× n^= ⁿ⁺¹₂

(c) E(X²) =^_n¹ × 1^2+^_n¹ × 2^2+ . . . +^1_n× n^2= (n+1)(2n+1) 6

Therefore,

Var(X) = E(X²) − (E(X))² = (n+1)(2n+1)

6 −^n+1₂ ^2 = ⁿ²₁₂⁻¹.

2.8.14 (a) Let XT be the amount of time that Tom spends on the bus and let XN be the amount of time that Nancy spends on the bus.

Therefore, the sum of the times is X = X_T + X_N and E(X) = E(X_T) + E(X_N) = 87 + 87 = 174 minutes.

If Tom and Nancy ride on different buses then the random variables XT and X_N are independent so that

Var(X) = Var(X_T) + Var(X_N) = 3²+ 3² = 18 and the standard deviation is√

18 = 4.24 minutes.

(b) If Tom and Nancy ride together on the same bus then X_T = X_N so that X = 2 × XT, twice the time of the ride.

In this case

E(X) = 2 × E(XT) = 2 × 87 = 174 minutes and

Var(X) = 2²× Var(X₁) = 2²× 3² = 36 so that the standard deviation is√

36 = 6 minutes.

2.8.15 (a) Two heads gives a total score of 20.

One head and one tail gives a total score of 30.

Two tails gives a total score of 40.

Therefore, the state space is {20, 30, 40}.

(b) P (X = 20) = ¹₄ P (X = 30) = ¹₂ P (X = 40) = ¹₄ (c) P (X ≤ 20) = ¹₄ P (X ≤ 30) = ³₄ P (X ≤ 40) = 1

(d) E(X) =^20 ×¹₄^+^30 ×¹₂^+^40 × ¹₄^= 30

(e) E(X²) =^20²×¹₄^+^30²× ¹₂^+^40²×¹₄^= 950 Var(X) = 950 − 30² = 50

2.8. SUPPLEMENTARY PROBLEMS 91

The standard deviation is√

50 = 7.07.

2.8.16 (a) Since R6

5 Ax dx = ^A₂ × (6²− 5²) = 1 it follows that A = ₁₁² .

(b) F (x) =^R₅^{x 2y}₁₁ dy = ^x2₁₁⁻²⁵

(c) E(X) =^R₅^{6 2x}₁₁² dx = ^2×(6₃₃³⁻⁵³⁾ = ¹⁸²₃₃ = 5.52

(d) E(X²) =^R₅^{6 2x}₁₁³ dx = ⁶⁴₂₂⁻⁵⁴ = ⁶⁷¹₂₂ = 30.5 Var(X) = 30.5 −^182₃₃^2 = 0.083

The standard deviation is√

0.083 = 0.29.

2.8.17 (a) The expectation is 3 × 438 = 1314.

The standard deviation is√

3 × 4 = 6.93.

(b) The expectation is 438.

The standard deviation is ^√4

8 = 1.41.

2.8.18 (a) If a 1 is obtained from the die the net winnings are (3 × $1) − $5 = −$2 If a 2 is obtained from the die the net winnings are $2 − $5 = −$3 If a 3 is obtained from the die the net winnings are (3 × $3) − $5 = $4 If a 4 is obtained from the die the net winnings are $4 − $5 = −$1 If a 5 is obtained from the die the net winnings are (3 × $5) − $5 = $10 If a 6 is obtained from the die the net winnings are $6 − $5 = $1 Each of these values has a probability of ¹₆.

(b) E(X) =^−3 × ¹₆^+^−2 × ¹₆^+^−1 ×¹₆^ + ^1 ×¹₆^+^4 ×¹₆^+^10 × ¹₆^= ³₂

E(X²) =^(−3)²× ¹₆^+^(−2)²×¹₆^+^(−1)²× ¹₆^ + ^1²× ¹₆^+^4²× ¹₆^+^10²×¹₆^= ¹³¹₆

The variance is

131

6 −^3₂^2 = ²³⁵₁₂

and the standard deviation is^q235₁₂ = $4.43.

(c) The expectation is 10 ׳₂ = $15.

The standard deviation is√

10 × 4.43 = $13.99.

2.8.19 (a) False (b) True (c) True (d) True (e) True (f) False

2.8.20 E(total time) = 5 × µ = 5 × 22 = 110 minutes The standard deviation of the total time is √

5σ =√

5 × 1.8 = 4.02 minutes.

E(average time) = µ = 22 minutes

The standard deviation of the average time is ^√σ

5 = ^√1.8

5 = 0.80 minutes.

2.8.21 (a) E(X) = (0 × 0.12) + (1 × 0.43) + (2 × 0.28) + (3 × 0.17) = 1.50 (b) E(X²) = (0²× 0.12) + (1²× 0.43) + (2²× 0.28) + (3²× 0.17) = 3.08

The variance is 3.08 − 1.50² = 0.83 and the standard deviation is√

0.83 = 0.911.

(c) E(X1+ X2) = E(X1) + E(X2) = 1.50 + 1.50 = 3.00 Var(X1+ X2) = Var(X1) + Var(X2) = 0.83 + 0.83 = 1.66 The standard deviation is√

1.66 = 1.288.

2.8.22 (a) E(X) = (−22 × 0.3) + (3 × 0.2) + (19 × 0.1) + (23 × 0.4) = 5.1

(b) E(X²) = ((−22)²× 0.3) + (3²× 0.2) + (19²× 0.1) + (23²× 0.4) = 394.7 Var(X) = 394.7 − 5.1²= 368.69

The standard deviation is√

368.69 = 19.2.

2.8.23 (a) Since R4

2 f (x) dx =^R₂⁴ _x^A2 dx = ^A₄ = 1 it follows that A = 4.

2.8. SUPPLEMENTARY PROBLEMS 93

(b) Since

1

4 =^R₂^yf (x) dx =^R₂^y _x⁴2 dx =^2 −⁴_y^ it follows that y = ¹⁶₇ = 2.29.

2.8.24 (a) 100 = E(Y ) = c + dE(X) = c + (d × 250) 1 = Var(Y ) = d²Var(X) = d²× 16

Solving these equations gives d = ¹₄ and c = ⁷⁵₂ or d = −¹₄ and c = ³²⁵₂ .

(b) The mean is 10 × 250 = 1000.

The standard deviation is√

10 × 4 = 12.65.

2.8.25 Since

E(c₁X₁+ c₂X₂) = c₁E(X₁) + c₂E(X₂) = (c₁+ c₂) × 100 = 100 it is necessary that c₁+ c₂= 1.

Also,

Var(c₁X₁+ c₂X₂) = c²₁Var(X₁) + c²₂Var(X₂) = (c²₁× 144) + (c²₂× 169) = 100.

Solving these two equations gives c₁= 0.807 and c₂ = 0.193 or c₁= 0.273 and c₂ = 0.727.

2.8.26 (a) The mean is 3µ_A= 3 × 134.9 = 404.7.

The standard deviation is√

3 σ_A=√

3 × 0.7 = 1.21.

(b) The mean is 2µA+ 2µB= (2 × 134.9) + (2 × 138.2) = 546.2.

The standard deviation is√

0.7²+ 0.7²+ 1.1²+ 1.1² = 1.84.

(c) The mean is ^4µA+3µ₇ ^B = (4×134.9)+(3×138.2)

7 = 136.3.

The standard deviation is

√0.7²+0.7²+0.7²+0.7²+1.1²+1.1²+1.1²

7 = 0.34.

Chapter 3

In document StatsSolutionManual (Page 50-96)