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The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 63 and a standard deviation of 10. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 63 and 83

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Answer:

47.5% of lightbulb replacement requests numbering between 63 and 83

Explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 63

Standard deviation = 10

What is the approximate percentage of lightbulb replacement requests numbering between 63 and 83

63 is the mean

83 = 63 + 2*20

So 83 is two standard deviations above the mean.

The normal distribution is symmetric, so 50% of the measures are above the mean and 50% below the mean.

Of those above the mean, 95% are within 2 standard deviations of the mean.

So

0.5*95% = 47.5%

47.5% of lightbulb replacement requests numbering between 63 and 83

User StrikeR
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