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.
Answer:
y =36
z= 54
Step-by-step explanation:
The angles 3y and 2y form a straight line, so they add to 180 degrees.
3y+2y =180
5y = 180
Divide each side by 5
5y/5= 180/5
y =36
The angles z, y, from a 90 degree angle
z+y =90
z+36 = 90
Subtract 36 from each side
z+36-36 = 90-36
z = 54
Answer:
c?
Step-by-step explanation: