A factory manager collected data on a number of equipment breakdowns per day. Fr
ID: 3024204 • Letter: A
Question
A factory manager collected data on a number of equipment breakdowns per day. From those data, she derived the probability distribution shown in the following table, where W denotes the number of breakdowns on a given day.
The factory manager estimates that each breakdown costs the company $500 in repair and loss of production. If W is the number of breakdowns in a day, then $500W is the cost of breakdowns for that day.
a.) Refer to the probability distribution table and determine the probability distribution of the random variable 500W.
b.) Determine the mean daily breakdown cost, 500W, by using your answer from part (a).
c.) Find 500W by using your answer from part (a).
I am very confused as to what a random variable is, as well as knowing how to solve the mean and S.D.
w 0 1 2 P(W=w) 0.80 0.15 0.05Explanation / Answer
Given the probability distribution of number of equipment breakdowns per day.
the cost of a breakdown is =$500
The random variable $500W takes values 0, $500, $1000 with probabilities 0.8,0.15 and 0.05 respectively.
Mean daily break down cost = sum( 500w* probability) = $125 (see below table for calculations)
standard deviationof daily break down cost 500W= sqrt( variacne of daily break down cost)
variance of daily break down cost = 87500 -(125)^2 =71875
standard deviationof daily break down cost 500W=sqrt(71875) = 268.095
w 0 1 2 500w 0 500 1000 P(w) 0.8 0.15 0.05Related Questions
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