Likewise
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XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:
Number of
Breakdowns------------- Probability
0------------------------------- 0.3
1 -------------------------------0.4
2------------------------------- 0.1
3------------------------------- 0.2
More than 3----------------- 0.0
(a) What is the expected number of breakdowns in any given day?
(b) What is the variance for this distribution?
(c) What is the probability that there will be at least 2 breakdowns in a day?
XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:
Number of
Breakdowns------------- Probability
0------------------------------- 0.3
1 -------------------------------0.4
2------------------------------- 0.1
3------------------------------- 0.2
More than 3----------------- 0.0
(a) What is the expected number of breakdowns in any given day?
(b) What is the variance for this distribution?
(c) What is the probability that there will be at least 2 breakdowns in a day?
Answers:
a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
A. You should know by definition why this isn't right. Expected value isn't (necessarily) a percentage, unless the random value is a percentage (but then it still wouldn't mean what it normally means. . .). The expected value is just that the value you would expect to get. In this case, since our random variable is a discreet number, referring to number of breakdowns, then we should get a specific number (and not necessarily an integer). To calculate expected value we take the sum of all random variables, multiplied by their probabilities: 0*P(X=0)+1P(X=1)+2P(X=2)+3P(X=3). Why does this work? Remember what the expected value is - the mean. Imagine you have a perfect sample of 100 days. Among this sample I would expect to have 30% have 0 breakdowns, 40% have 1 breakdown, 10% have 2 breakdowns, 20% have 3 breakdowns, and 0% to have more than 3 breakdowns. That means I would EXPECT to have, among a perfect sample of 100: 30 0's, 40 1's, 10 2's, 20 3's and 0' "More than 3's". Now how would I go about averaging those numbers? And that is your expected value.