Question

A light bulb manufacturer considers a produced light bulb good if its luminance falls within 3 SD fromthe desired value. Suppose the expected value of each luminance measurement equals the desired value.What percent of items will be considered good, if the distribution of measurements is(a) Normal(μ,σ2)(b) Uniform(a,b)?

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Answer 1

(a)

For Normal Distribution:

By Empirical Rule:

Percent of items that will be good.i.e., 3 SD from the desired value = 99.7 %

So,

Answer is:

99.7 %

(b)'

For Uniform Distribution (a,b):

Mean =

.

SD is given by:

.

So,

Mean +3 SD is given by:

Mean - 3 SD is given by:

Further, we note:

.

Thus, we note the interval: Mean - 3 SD to Mean + 3 SD is wider than the entire interval of the given Uniform Distribution (a,b).

So,

we get :

The probability of Mean - 3 SD to Mean + 3 SD is 1 in the case of Uniform Distribution.

So,

Answer is:

100 %