Problem 1. See section 5.2 of the text and the exercises for this section on pag

Question

Problem 1. See section 5.2 of the text and the exercises for this section on page 204. For the past four years, Armonco Manufacturing has been offering a three-year limited warranty on all appliances it manufactures. Although all appliances are given a unique serial number when manufactured, until this year Armonco had no capability of determining how often any appliance was brought to an authorized service facility. At the beginning of the year, the long-promised computer database linking all service facilities with a central system was finally operational. A preliminary report shows the following results for one of the appliances Armonco manufactures. Times Brought for Repair Probability 0 0.55 1 0.25 2 0.14 3 0.04 4 0.02 a) Find the expected number of repairs for this appliance (distribution mean). b) Find the standard deviation of the repair distribution. Problem 2. Refer to section 5.3 of text. For a sample of n=10, assuming that the binomial probability distribution applies, find the following. a) the probability of 3 successes if the probability of a success is 0.20 b) the probability of 3 successes if the probability of a success is 0.80 Problem 3. Refer to section 5.4 of text. Your company president has told you that the company experiences product returns at the rate of two per month with the number of product returns distributed as a Poisson reandom variable. Determine the probability that next month there will be: a) no returns. c) two returns. d) What is the probability that the company will experience three returns in the next two months.

Explanation / Answer

a)Expected value of repairs= summation ( no. of times brought for repair* probability) = 0*0.55+1*0.25+2*0.14+3*0.04+4*0.02 =0.25+0.28+0.12+0.08 = 0.73 so the expected number of times is 0.73 b) variance = summation( (N-mean)^2 *probability) ) = 0.7696 so standard deviation = sqrt ( variance ) = 0.877 Problem 2: a) Here as the distribution is binomial, so probability is the coefficient of third term P = 10C3 x 0.2^3 x 0.8 ^7 = 0.20133 b) P= 10C3 x 0.8^3 x 0.2 ^7 = 7.864*10^-4

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