The maintenance department at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 43 and a standard deviation of 11. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 10 and 43

The approximate percentage of lightbulb replacement requests numbering between 10 and 43 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 = 43

Standard deviation = 11

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

What is the approximate percentage of lightbulb replacement requests numbering between 10 and 43?

43 is the mean

10 = 43 - 3*11

So 10 is 3 standard deviations below the mean.

Of the 50% of measures below the mean, 99.7% are between 3 standard deviations below the mean(10) and the mean(43). So

0.5*0.997 = 0.4985.

0.4985*100% = 49.85%.

The approximate percentage of lightbulb replacement requests numbering between 10 and 43 is of 49.85%.

its b

step-by-step explanation:

answer: im gonna go with milli

step-by-step explanation: