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The probability distribution for the number of defects during an eight hour shift on the assembly line at Wanda’s Wooden Widgets is as shown in the chart below.x 0 1 2 3 4 5P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01On average, how many defects are found during an 8-hour shift? 5.3 2.5 0.9 0.50 0.1667

Question

The probability distribution for the number of defects during an eight hour shift on the assembly line at Wanda’s Wooden Widgets is as shown in the chart below.x 0 1 2 3 4 5P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01On average, how many defects are found during an 8-hour shift? 5.3 2.5 0.9 0.50 0.1667

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Solution

To find the average number of defects found during an 8-hour shift, we need to calculate the expected value of the probability distribution. The expected value (E) is calculated by multiplying each outcome by its probability and then summing these products.

Here's how to do it:

E = (0 * 0.50) + (1 * 0.25) + (2 * 0.15) + (3 * 0.06) + (4 * 0.03) + (5 * 0.01)

E = 0 + 0.25 + 0.30 + 0.18 + 0.12 + 0.05

E = 0.90

So, on average, 0.9 defects are found during an 8-hour shift.

This problem has been solved

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