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.


Now we will find z-score corresponding to 56.

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
.
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.

Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
3w-13w=-20
Combine like terms
-10w=-20
divide both sides by -10
w=2
Answer: what missing exponent?
Step-by-step explanation:
The confidence interval is

We first find p, our sample proportion. 118/200 = 0.59.
Next we find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1: 1-0.98 = 0.02
Divide by 2: 0.02/2 = 0.01
Subtract from 1: 1-0.01 = 0.99
Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 2.33.
The margin of error (ME) is given by

This gives us the confidence interval