A. Suppose for a particular shipment of devices, it is reasonable assume the num

Question

A. Suppose for a particular shipment of devices, it is reasonable assume the number of defective devices follows a Binomial distribution. Assume a shipment consists of 400 devices and that the probability a device is defective is 0.03. What is the mean and standard deviation (in that order) for X, the number of devices in the shipment with a defect?

B. Suppose X represents whether or not a randomly selected device is non-defective. It takes on the value 1 if the device is non-defective and 0 if it is not. Does this seem more like the Binomial setting, the Poisson setting, or the Normal setting?

Explanation / Answer

A. n = 400 p = 0.03

1 - p = 1 - 0.03 = 0.97

Mean = np = 400 * 0.03 = 12.

Standard deviation = (np(1-p)) = (12*0.97) = 3.4117.

B. Since X can take only two values, this represents a Bernoulli trial or the binomial setting.

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