The Garden Company Ltd. produces a special brand of cake, sold in “500g” packets. The
weights of these packets are normally distributed with a mean of 520g and a standard
deviation of 12g.
a) What is the probability that a randomly selected packet weighs less than 500g?
(4 marks)
b) Out of a batch of 100 packets, how many would you expect to weigh less than 500g
(to the nearest whole number)? (2 marks)
c) It has been decided to reduce the mean weight to 510g. What should the standard
deviation be if only 1% of packets should weigh less than 500g? (4 marks)
II. Four watches are subjected to similar breaking tests independently. The probability that a
watch will be broken during the test is 0.40.
a) Write down the probability distribution for the number of watches that fail the
breaking tests. Give any TWO reasons to support your claim. (3 marks)
b) If four watches are tested on one particular occasion,
i) what is the probability that two or more watches will be broken? (3.5 marks)
ii) what is the probability that at most one watch will be broken? (2 marks)
iii) how many watched are expected to be broken during the test?
Probabitily
2011-01-03 12:34 am
回答 (2)
2011-01-03 2:31 am
✔ 最佳答案
1. (a) Denote X = packet weight; X ~ N(520,12^2=144)P(X < 500) = P(Z = (X - μ)/σ < (500-520)/12 = -5/3) (Z ~ N(0,1))= P(Z < -5/3)= Φ(-5/3) (Φ: cdf of N(0,1))= 1 - Φ(5/3)= 1 – (0.9525) (From the standard normal table[1])= 0.0475 (b) Expected no. of packets = 100*P(X < 500) = 100*(0.475) = 4.75 ~ 5 (c) Let the new distribution be N(510,σ2). P(X < 500) = 1% = 0.01 P(Z = (X - μ)/σ < (500-510)/σ = -10/σ) = 0.01 P(Z < -10/σ) = Φ(-10/σ) = 0.011 - Φ(10/σ) = 0.01Φ(10/σ) = 0.9910/σ = 2.33 (From the standard normal table[1])σ = 4.29 2. (a) Denote Y = No. of watches will be broken during the test. Since each test is independent to one another and has only two possible outcomes, broken or not (success or
failure), with equal probability p = 0.40 to result in a
broken watch, the tests are independently and identically
distributed as Bernoulli(0.40). Therefore, Y~ Binomial(4,0.40).
(b) (i) P(Y ≧ 2) = 1 – P(Y < 2) = 1 - 1Σk=0 4Ck(0.40)k(1 – 0.40)4 - k= 1 – 0.4752= 0.5248 (ii) P(Y ≦ 1) = 1 - P(Y > 1) = 1 - P(Y ≧ 2) = 1 – (0.5248) = 0.4752 (iii) Expected no. of broken watches: μY = E(Y) = (4)*(0.40) = 1.6
參考: myself, [1] the standard normal table which is available at: http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm
2011-01-03 1:03 am
ProbabitilyI (a) P(X<500)
=P(Z< (500-520)/12)
=P(Z<-1.67)
=0.04746
(b) 100*0.04746=5 (to the nearest whole number)(c) P(X<500) = 0.01
P(Z< (500-mu)/sigma) = 0.01
(500-mu)/sigma = -2.3263
sigma=4.2987II(a) X~Bin(4,0.4) where X is the number of watches which are broken during the test. Reason: 1 Each test has only two outcomes. 2 Tests are independent.(b)(i) 1-P(0)-P(1)=1-(0.6)^4-4(0.4)(0.6)^3=0.5248(ii) P(0)+P(1)=(0.6)^4+4(0.4)(0.6)^3=0.4752(iii) Expectation=np=4*0.4=1.6
=P(Z< (500-520)/12)
=P(Z<-1.67)
=0.04746
(b) 100*0.04746=5 (to the nearest whole number)(c) P(X<500) = 0.01
P(Z< (500-mu)/sigma) = 0.01
(500-mu)/sigma = -2.3263
sigma=4.2987II(a) X~Bin(4,0.4) where X is the number of watches which are broken during the test. Reason: 1 Each test has only two outcomes. 2 Tests are independent.(b)(i) 1-P(0)-P(1)=1-(0.6)^4-4(0.4)(0.6)^3=0.5248(ii) P(0)+P(1)=(0.6)^4+4(0.4)(0.6)^3=0.4752(iii) Expectation=np=4*0.4=1.6
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