In this recession, yours truly, CEO of the Outrageous Products Enterprise, would like to make extra money to support my frequent filet-mignon-and-double-lobster-tail dinner habit. A promising enterprise is to mass-produce tourmaline wedding rings for brides. Based on my diligent research, I have found out that women's ring size normally distributed with a mean of 6.0, and a standard deviation of 1.0. I am going to order 5000 tourmaline wedding rings from my reliable Siberian source. They will manufacture ring size from 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, and 9.5. How many wedding rings should I order for each of the ring size should I order 5000 rings altogether?


 what we have to do is notice that the ring sizes or 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9 and 9.5 are a certain number of standard deviations from the mean. This is how we will solve this
4 is -2 st dev obtained by taking (4 - 6)/1 = -2
4.5 is -1.5 st dev by taking (4.5 - 6)/1 = -1.5 etc
5 is -1 st dev
5.5 is -0.5 st dev
6 is the mean , so 0 st dev
6.5 is 0.5 st dev
7 is 1 st dev
7.5 is 1.5 st dev
8 is 2 st dev
8.5 is 2.5 st dev
9 is 3 st dev
9.5 is 3.5 st dev.
This is how we will figure out what percentage of each ring should be bought. We can't just say we look at standard normal distribution probabilities and take P(Z = -2) for size 4.0 because remember in a continuous distribution, there are no probabilities for exact values. So we take P(-2.5 < Z < -2.0) = .0166 (i did half standard deviations because each ring size is half st dev apart) . Now take .0166(5000) = 83
The same method is applied to each size
size 4.5 = P(-2 < Z < -1.5) = .044, multiplied by 5000 = 220
size 5 = P(-1.5 < Z < -1) = .0919, multiplied by 5000 = 460
size 5.5 = P(-1 < Z < -0.5) = .1498, multiplied by 5000 = 749
size 6 = P(-0.5 < Z < 0) = .1915, multiplied by 5000 = 958
Since this is the normal distribution, this is symmetrical, so size 6.5 = 958, size 7 = 749, size 7.5 = 460, size 8 = 220, size 8.5 = 83
For size 9, same idea = P(2.5 < Z < 3) = .0049 multiplied by 5000 = 25
size p.5 = P(3 < Z < 3.5) = .0011 times 5000 = 6
Now that gives a total of 4971, the rest are either bigger than 9.5 or less than 4, since no rings will be produced bigger than size 9.5, the other 29 should be of size 4, making size 4 total 112