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:
C
Step-by-step explanation:
Add all the schools together, then divide by 5
5805/5
Ans: 1.1 miles more
Speed = distance/time
Time is 1/5 of an hour
1/5 of 1hr = 1/5 or 0.2 hrs
Speed = 16 mph
So, we're after distance.
Distance = speed x time
= 16mph x 0.2
= 3.2 miles
Walks for 3/7 hour at 5 miles per hour :
Time & speed given, distance is what we're after.
(3/7 of 1hr is 3/7)
Distance = speed x time
= 5mph x 3/7
= 15/7 miles or 2.14285...miles
So, distance is about 2.1 miles to 1 d.p
Thus, 3.2 - 2.142... (the full ans) = 1.057... miles
Well, he biked 1.06 miles or 1.1 miles (depending on how you want to round it) more than walking.
Hope this helps!