Probability help!!!!!!!!!!!!!!

[匿名]

2013-10-12 6:24 am

Discrete ProbabilityDistribut
4.Based on reent records, the manager of a car painting center has determined the following probability distribution for the number of customers per day. The center has the capacity to serve two customers per day.

xp(x)
00.05
10.20
20.30
30.25
40.15
50.05

(a)What is the probability that one or more customers will be turned away on a given day?

(b)What is the probability that the center’s capacity will not be fully utilized on a day?

(c)By how much must the capacity be increased so the probability of tuning a customer away is no more than 0.1?

5.A salesman of home computers will contact four customers during a week. Each contact can result in either a sale, with probability 0.2, or no sale, with probability 0.8. Assume that customer contacts are independent.

(a)List the elementary outcomes and assign probabilities.

(b)If X denotes the number of computers sold during the week, obtain the probability distribution of X.

(c)Calculate the expected value and the standard deviation of X.

回答 (1)

myisland8132

2013-10-12 11:12 pm

✔ 最佳答案

4(a) probability that one or more customers will be turned away on a given day

= 0.25 + 0.15 + 0.05

= 0.45

(b) probability that the center’s capacity will not be fully utilized on a day

= 0.05 + 0.2

= 0.25

(c) The capacity should be increased by 4 - 2 = 2

5(a) elementary outcomes: sell a computer, do not sell a computer

with probabilities 0.2 and 0.8 respectively

(b) X = 0 ,P(0) = 0.8^4 = 0.4096

X = 1,P(1) = C(4,1) * 0.8^3 * 0.2 = 0.4096

X = 2,P(2) = C(4,2) * 0.8^2 * 0.2^2 = 0.1536

X = 3,P(3) = C(4,3) * 0.8^3 * 0.2 = 0.0256

X = 4, P(4) = 0.2^4 = 0.0016

(c) expected value = np = 4 * 0.2 = 0.8

standard deviation of X = (npq)^0.5 = (4 * 0.2 * 0.8)^(0.5) = 0.8

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