Standard Deviation: average worker at the call center answers an average of 11 calls

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smithy38

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Am I doing this Correctly;

The average worker at the call center answers an average of 11 calls per hour with a population standard deviation of 3 calls/hour.
What is the probability that a worker will answer more than 15 calls in an hour?
How I approached solving the problem:
The mean is 11
Standard deviation is 3 (this confused me since it said population standard deviation)
Z score = Measured value – mean / standard deviation
Z score = (15 – 11)/3 = 1.33 or 0.9082.
If we are looking at more than this becomes the value to the right of the curve or 1- 0.9082 = 9.18%
Answer:
9.18% is the probability that a worker will answer more than 15 calls in an hour

What is the probability that a worker will answer between 7 and 12 calls in an hour?
How I approached solving the problem:
Z score = (7 – 11)/3 = -1.3333 or
Z score = (12 – 11)/3 = .3333
We want the values in between the two scores (between 0.0919 and 0.6293); this equals 0.6293 -.0919.

Answer: is 0.5374 or 53.74% probability that a worker will answer between 7 and 12 calls in an hour.
 
smithy38 said:
Am I doing this Correctly;

The average worker at the call center answers an average of 11 calls per hour with a population standard deviation of 3 calls/hour.
What is the probability that a worker will answer more than 15 calls in an hour?
How I approached solving the problem:
The mean is 11
Standard deviation is 3 (this confused me since it said population standard deviation)
Z score = Measured value – mean / standard deviation
Z score = (15 – 11)/3 = 1.33 or 0.9082.
If we are looking at more than this becomes the value to the right of the curve or 1- 0.9082 = 9.18%
Answer:
9.18% is the probability that a worker will answer more than 15 calls in an hour

What is the probability that a worker will answer between 7 and 12 calls in an hour?
How I approached solving the problem:
Z score = (7 – 11)/3 = -1.3333 or
Z score = (12 – 11)/3 = .3333
We want the values in between the two scores (between 0.0919 and 0.6293); this equals 0.6293 -.0919.

Answer: is 0.5374 or 53.74% probability that a worker will answer between 7 and 12 calls in an hour.
Click to expand...

Have you learnt about adjusting for discrete values? Note that the variable here is the number of calls (which must be an integer value and therefore discrete).
 
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