We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.

First of all, we will find z-score corresponding to 38 and 56.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{38-38}{6}=\frac{0}{6}=0$

Now we will find z-score corresponding to 56.

$z=\frac{56-38}{6}=\frac{18}{6}=3$

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is $-3\sigma\text{ to }3\sigma$ .

We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.

We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

$\frac{99.7\%}{2}=49.85\%$

Therefore, approximately $49.85\%$ of lightbulb replacement requests numbering between 38 and 56.

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3 years ago