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Solutions to Exercises 8
- (a) Assuming that calls are answered independently, X ∼ Bin(20, 0 .85). (b) The mean and variance are E(X) = np = 20 × 0 .85 = 17 V ar(X) = np(1 − p) = 20 × 0. 85 × 0 .15 = 2. 55
and so
SD(X) =
V ar(X) =
(c) P (X = 9) = nCrpr(1 − p)n−r = 20 C 9 × 0. 859 × (1 − 0 .85)^20 −^9 = 20! 9! × 11!
× 0. 859 × 0. 1511
=^20 ×^19 × · · · ×^12
9 × 8 × · · · × 1
× 0. 859 × 0. 1511
= 167960 × 0. 859 × 0. 1511
(d) P (X < 2) = P (X = 0) + P (X = 1) = 20 C 0 × 0. 850 × (1 − 0 .85)^20 −^0 + 20 C 1 × 0. 851 × (1 − 0 .85)^20 −^1 = 0. 1520 + 20 × 0. 85 × 0. 1519 = 3. 3255 × 10 −^17 + 3. 76862 × 10 −^15 = 3. 802 × 10 −^15.
- (a) X ∼ P o(10) (b) The mean and variance are E(X) = λ = 10 V ar(X) = λ = 10
and so
SD(X) =
V ar(X) =
(c)
P (X = 12) = λ
re−λ r! =^10
(^12) × e− 10 12! = 0. 09478.
(d)
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
=^10
(^0) × e− 10 0! +
101 × e−^10 1! +
102 × e−^10 2! = e−^10 + 10e−^10 + 50e−^10 = 0.0000454 + 0.0004540 + 0. 0022700 = 0. 00277.
- Let Y be the total sales in the 5-day period.
(a) E(Y ) = 5 × 2 = 10. So Y ∼ Po(10). (b)
P (Y > 12) = 1 − P (Y ≤ 12) = 1 − {P (Y = 0) + P (Y = 1) + · · · + P (Y = 12)} = 1 −
e−^10100 0! +^
e−^10101 1! +^ · · ·^ +^
e−^101012 12!
= 1 − e−^10
2! +^ · · ·^ +
= 1 − e−^10 {1 + 10 + 50 + · · · + 2087. 6757 }
(The first term (Y = 0) is 1. The second (Y = 1) is 10 / 1 times the first. The third (Y = 2) is 10 / 2 times the second and so on. Alternatively you can use Minitab).
P (Y > 12) = 1 − e−^10 { 17435. 1965 } = 1 − 0. 7916 = 0. 2084.
- You might find a probability tree helpful in answering this question.
(a)
P (compt accptd |machine OK) = P (compt OK and accptd |machine OK) +P (compt defective and accptd |machine OK) = 0. 95 × 0 .97 + 0. 05 × 0. 15 = 0. 929.
(b)
P (compt rejected |machine OK) = 1 − P (compt accptd |machine OK) = 1 − 0. 929 = 0. 071.
(j)
P (machine OK |2 accptd out of 5) =
P (machine OK and 2 accptd out of 5) P (2 accptd out of 5) = 0.^0027800
- 0075232 = 0. 3695
Notice that, having seen the evidence of two components accepted out of five, the probability that the machine requires adjustment changes from 0. 1 to 1 − 0 .3695 =
