Need help with this problem. Can someone please solve ?
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)14.22 pounds. b)15.08 pounds. c)14.00 pounds.
z 0.04 0.05 1.8 0.9671 0.9678 1.9 0.9738 0.9744 2.0 0.9793 0.9798 2.1 0.9838 0.9842
First you need to find the z-value that corresponds to 98% of the distribution. It looks like at a z-value of 2.05 , 98% of your distribution lies to the left, so the “top 2%” must lie to the right. This is where your “heavy” bowling balls should be, 2.05 standard deviations away from the mean.
If your standard deviation is 1.5 lbs, and the top 2% (heavy) lie 2.05 standard deviations away from your mean, then: 1.5 lbs x 2.05 = 3.075 lbs heavier than your average (mean) bowling ball is where the “heavy” category should start. 12 + 3.075 = 15.075 or 15.08 rounded.
Hope this was clear enough!
I will provide a shortcut (six of one, half a dozen of another). Since they want you to use the z-distribution, you shouldn_'t need_ the tables (I believe common critical values are fair game to give without a table), especially since it’s unlikely they’ll give you answers that are too close to call without a calculator.
You should know that -1.96 represents the 2.5th percentile and +1.96 represents the 97.5th percentile for this distribution. so 97.5 is really close to 98th percentile, and 1.96 is really close to 2. Multiply the SD by 2 to arrive at 3, and add 3 to 12. A weight of 15 is approximately at the 98th percentile of the distribution.
In your head it should be: z distribution, 1.96 is 97.5th percentile ~ 98th, calculation of 2 is close enough and easier to use, 1.5*2 + 12 = 15 so about 15 puts us at the 98th percentile.
Thanks for the help guys. Appreciate it. 