Question

The maintenance department at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 42 and a standard deviation of 10. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 42 and 62.

Answer 1

we are given

mean=42

Std=10

if the mean=42 + std =10 42+10=52

if the mean=42 - std=10 42-10=32

Rule -- 68-95-99.7

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

42+10=52

95% are within 2.

42+20=62

99.7% are within 3.

42+30=72

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

we are ask for the porcentage of request between 42-62 (between the mean and 2+std)

62 is two standard deviations above the mean.

Of the 50% of the measures below the mean, 95% are between 42 and 62, so

0.95(50)=47.5

The approximate percentage of light bulb replacement requests numbering between 42 and 62 is of 47.5%