The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 47 and a standard deviation of 6. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 47 and 65

The approximate percentage of lightbulb replacement requests numbering between 47 and 65 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 of 47, standard deviation of 6.

What is the approximate percentage of lightbulb replacement requests numbering between 47 and 65?

65 = 47 + 3*6

So 65 is three standard deviations above the mean, and this percentage is the percentage between the mean and 3 standard deviations above the mean.

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

Of those 50% above, 99.7% are within 3 standard deviations of the mean, so:

0.997*0.5 = 0.4985.

0.4985*100% = 49.85%.

The approximate percentage of lightbulb replacement requests numbering between 47 and 65 is of 49.85%.

the answer is 62

the first one

step-by-step explanation:

just id the test.