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

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