# Quant Help

The owner of a bowling alley determined that the average weight for a bowling ball is 12 pounds with a standard deviation of 1.5 pounds. A ball denoted “heavy” should be one of the top 2% based on weight. Assuming the weights of bowling balls are normally distributed, at what weight (in pounds) should the “heavy” designation be used? A) 14.00 pounds. B) 14.22 pounds. C) 17.36 pounds. D) 15.08 pounds.

is it D?

Yes. Can you provide an explaination? I fell for B.

So a heavy ball is in the top 2%, that means to the right you have 2%, to the left you have 98%. Check a cummulative probability distribution, z for 98% is somewhere between 2.05 and 2.06, take an average of 2.055. You have the average 12+2.055*1.5=15.0825.

Being I thought this was a confidence interval question, can you tell me where you found the the Z value?

A-66 in Volume I of CFAI text.

Another quick way to eliminate some answers would be noting that if the distribution is assumed normal, you know that the top 2% means that 98% are less than it. You know that to get up to 95% of values is 2 deviations from the mean, so 1.5+1.5+12=15lbs. You should know then that the answer has to be over 15 to be in the top 5%… While this doesnt give you the answer, it sure narrows it down quickly, hope it helps

14-12/1.5 = 1.333 < 1.645 so less than 68% 14.22 - 12 / 1.5 = 1.48 < 68% C) 17.36 pounds. (17.36 - 12) / 1.5 = 3.573, assume normal > 99% because 2.58 is the threshold. D) 15.08 pounds. 15.08 - 12 / 1.5 = 2.053 so D must be the answer.

tvPM Wrote: ------------------------------------------------------- > Another quick way to eliminate some answers would > be noting that if the distribution is assumed > normal, you know that the top 2% means that 98% > are less than it. You know that to get up to 95% > of values is 2 deviations from the mean, so > 1.5+1.5+12=15lbs. You should know then that the > answer has to be over 15 to be in the top 5%… > > While this doesnt give you the answer, it sure > narrows it down quickly, hope it helps Remember that : for two tails, 68% are in +/- 1 stdv, leaving 16% in each tail for two tails, 90% are in +/-1.645 stdv, leaving 5% in each tail for two tails, 95% are in +/-1.96 stdv, leaving 2.5% in each tail, you can use this for approximation on 1 tail for 98% to the left, 2% to the right for two tails, 98% are in +/-2.58 stdv, leaving 1% in each tail for two tails, 99.7% are in +/- 3 stdv, leaving 0.15% in each tail CPK is more to the point.