Solutions to Assignment 1 (STA 4443)

5.2-2. (a) P (0 ≤ Z ≤ 0.87) = Φ(0.87) − Φ(0) = 0.8078 − 0.5 = 0.3078.(b) P (−2.64 ≤ Z ≤ 0) = Φ(0)−Φ(−2.64) = Φ(0)−[1−Φ(2.64)] = 0.5−(1−0.9959) =0.4959.(c)

P (−2.13 ≤ Z ≤ −0.56) = Φ(−0.56) − Φ(−2.13)

= [1 − Φ(0.56)] − [1 − Φ(2.13)] = Φ(2.13) − Φ(0.56)

= 0.9834 − 0.7123 = 0.2711.

(d)

P (|Z| > 1.39) = P (Z > 1.39) + P (Z < −1.39) = 2P (Z > 1.39)

= 2[1 − Φ(1.39)] = 2[1 − 0.9177] = 0.1646.

(e)P (Z < −1.62) = Φ(−1.62) = 1 − Φ(1.62) = 1 − 0.9474 = 0.0526.(f)P (|Z| > 1) = 2P (Z > 1) = 2[1 − Φ(1)] = 2[1 − 0.8413] = 0.3174.(g) P (|Z| > 2) = 2P (Z > 2) = 2[1 − Φ(2)] = 2[1 − 0.9772] = 0.0456.(h) P (|Z| > 3) = 2P (Z > 3) = 2[1 − Φ(3)] = 2[1 − 0.9987] = 0.0026.

5.2-6. By the given moment generating function M(t) = exp(166t+400t2/2), we findthat(a) µ = 166; (b) σ2 = 400.(c) P (170 < X < 200) = P (170−166

20< X−166

20< 200−166

20) = P (0.2 < Z < 1.7) = P (Z <

1.7) − P (Z ≤ 0.2) = 0.9554 − 0.5793 = 0.3761.(d) P (148 ≤ X ≤ 172) = P (−0.9 ≤≤ 0.3) = P (Z ≤ 0.3) − P (Z < −0.9) = P (Z ≤0.3) − P (Z > 0.9) = 0.6179 − 0.1841 = 0.4338.

5.2-14. (a) X has a normal distribution N(21.37, 0.16). So, µ = 21.37 and σ = 0.4.Now, we have

P (X > 22.07) = P

(

X − 21.37

0.4>

22.07 − 21.37

0.4

)

= P (Z > 1.75) = 1 − P (Z ≤ 1.75) = 1 − Φ(1.75)

= 1 − 0.9599 = 0.0401.

(b) We can easily see that Y has a binomial distribution b(n, p), where n = 15 andp is the probability that the weight of a randomly selected single mint is less than

1

20.857 grams. In order to find P (Y ≤ 2), we need to find p. We have

p = P (X < 20.857) = P

(

X − 21.37

0.4<

20.857 − 21.37

0.4

)

= P (Z < −1.2825) = Φ(−1.2825) = 1 − Φ(1.2825)

= 1 − 0.8997 = 0.1003.

So,

P (Y ≤ 2) = P (Y = 0) + P (Y = 1) + P (Y = 2)

=

(

15

0

)

(1 − 0.1003)15 +

(

15

1

)

(0.1003) · (1 − 0.1003)14

+

(

15

2

)

(0.1003)2 · (1 − 0.1003)13.

= 0.8159.

5.3-4. (a) P (X < 6.0171) = P (X−6.050.02

< 6.0171−6.050.02

) = P (Z < −1.645) = P (Z >1.645) = 0.05.(b) Let Y be the number of boxes that weigh less than 6.0171 pounds. Then Y isb(9, 0.05) and P (Y ≤ 2) = 0.9916 by using Table II.(c) P (X ≤ 6.035) = P ( X−6.05

0.02/3≤ 6.035−6.05

0.02/3) = P (Z ≤ −2.25) = P (Z ≥ 2.25) =

0.0122.

5.3-6. (a) Using χ2(16), we have

P (796.2 ≤∑16i=1(Xi − 50)2 ≤ 2630) = P (796.2

100≤P

16

i=1(Xi−50)2

100≤ 2630

100) = 0.95 − 0.05 =

0.90.(b) Using χ2(15), we have

P (726.1 ≤∑16i=1(Xi − X)2 ≤ 2500) = P (726.1

100≤P

16

i=1(Xi−X)2

100≤ 2500

100) = 0.95 − 0.05 =

0.90.

5.3-14. (a) E(X) = 24.5, V ar(X) = 3.82/8 = 1.805.E(Y ) = 21.3, V ar(Y ) = 2.72/8 = 0.911.(b) X − Y has approximately normal distribution N(24.5 − 21.3, 1.805 + 0.911) =N(3.2, 2.716).(c) P (X > Y ) = P (X − Y > 0) = P (Z > 0−3.2

√

2.716) = P (Z > −1.94) = P (Z < 1.94) =

0.9738.

2

5.4-2. If f(x) = (3/2)x2, −1 < x < 1, then we have

E(X) =

∫ 1

−1

x(3/2)x2dx = 0;

Var(X) =

∫ 1

−1

x2(3/2)x2dx =[

(3/10)x5]1

−1= 3/5.

Thus, we find

P (−0.3 ≤ Y ≤ 1.5) = P

(

−0.3 − 0√

15(3/5)≤ Y − 0√

15(3/5)≤ 1.5 − 0√

15(3/5)

)

≈ P (−0.10 ≤ Z ≤ 0.50) = P (Z ≤ 0.50) − P (Z ≤ −0.10)

= 0.6915 − 0.4602 = 0.2313.

5.4-6. (a)

µ =

∫ 2

0

x(1 − x/2)dx =

[

x2

2− x3

6

]2

0

= 2 − 4

3=

2

3.

σ2 =

∫ 2

0

x2(1 − x/2)dx −(

2

3

)2

=

[

x3

3− x4

8

]2

0

− 4

9=

2

9.

(b)

P

(

2

3≤ X ≤ 5

6

)

= P

(

2/3 − 2/3√

(2/9)/18≤ X − 2/3√

(2/9)/18≤ 5/6 − 2/3√

(2/9)/18

)

≈ P (0 ≤ Z ≤ 1.5) = P (Z ≤ 1.5) − P (Z ≤ 0)

= 0.9332 − 0.5000 = 0.4332.

5.4-12. Note that the distribution of X is N(2000, 5002/25). Thus, we have

P (X > 2050) = P

(

X − 2000

500/5>

2050 − 2000

500/5

)

≈ P (Z > 0.5) = 1 − P (Z ≤ 0.5) = 1 − 0.6915 = 0.3085.

5.4-18. We are given that the number of sick days Y =∑20

i=1 Xi has mean 200 andvariance 20 × 22 = 80. We want to find y (the budgeted days) so that

P (Y ≥ y) < 0.20.

3

It is equivalent to

P

(

Y − 200√80

>y − 200√

80

)

< 0.20.

By the central limit theorem, we find that approximately

y − 200√80

= 0.84,

which gives y = 207.51 and rounding to y = 208 days.

4