|
Question 1201185: In the claims department of an insurance office various quantities are
computed at the end of each day’s business. On Monday, 20 claims are
received for a particular class of policy. The mean claim amount is calculated
to be K4,500 and the standard deviation to be K2,540. On Tuesday, the
claims are reviewed and one claim which was incorrectly recorded as K13,000
is now corrected to K3,000. Determine the mean and standard deviation of
the corrected set of claims
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think it's going to work like this.
the mean of 4500 * 20 = sum of 90,000
subtract 13000 from that and add 3000 to that leaves a sum of 80000.
divide that by 20 to get a new mean of 4000.
square the original standard deviation of 2540 to get a variance of 64516001.
mutliply that by 20 to get a sum of squares of 129032000.
13000 - 4500 = original deviation of 8500.
square that to get a squared deviation of 72250000.
subtract that from the original sum of squares to get a sum of squares of 56782000.
3000 - 4000 = revised deviation of -1000.
square that to get a squared deviation of 1000000.
add that to the sum of squares to get a revised sum of squares of 57782000.
divide that by 20 to get a new variance of 2889100.
take the square root of that to get a new standard deviation of 1699.735274.
if i did this correctly, your solution should be:
mean of corrected claims is 4000.
standard deviation of corrected claims is 1699.735274.
here's a reference on how to calculate standard deviation.
https://www.mathsisfun.com/data/standard-deviation-formulas.html
|
|
|
| |
