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Katarina [22]
3 years ago
9

The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbul

bs. The distribution of the number of daily requests is bell-shaped and has a mean of 54 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?
Mathematics
1 answer:
amm18123 years ago
6 0

Answer:

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

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

Approximately 95% of the measures are within 2 standard deviations of the mean.

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

In this problem, we have that:

Mean = 54, standard deviation = 3.

What is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?

63 = 54 + 3*3

So between the mean and 3 standard deviations above the mean.

The normal distribution is symmetric, which means that 50% of the values are below the mean and 50% are above.

Of those 50% above, 99.7% are below 63. So

0.5*0.997 = 0.4985

0.4985*100% = 49.85%

The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.

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