laurjohns7 said:
Suppose that the weight of boxes of Oatsies ceral is normally distributed with a mean of 11 ounces and standard deviation of 2 ounces. In a shipment of 10,000 boxes of cereal
how many cuold be expected to weigh between 8 and 10 ounces
First, find the z-score by using \(\displaystyle z=\frac{(x-{\mu})}{\sigma}\)
You are given \(\displaystyle {\mu}=11, \;\ \sigma=2, \;\ n=10000\)
Plug in x=8 and x=10 into the formula and find the z-score for each. Look them up in the z-table, then subtract the probabilities.
Multiply this number by n=10,000 to find out how many fall in the interval 8 to 10.
weigh more than 18 ounces
Do the same as above except subtract the probability from 1. Then, multiply by 10000.
The reason we subtract from 1 is the z-scores come from the left and go right. For 'more than' a certain value we must then subtract from 1
because the total area under the normal distribution curve is 1.
For this one, since x=18, the mean is 11, and the standard deviation is 2, we can expect it to be a very low number compared to 10000.
See, 2 units right and left of the mean, 11, is 1 S.D. By the Empirical Rule, in the interval 9 to 13 ounces, about 68% (or 6800) of the boxes will be in this weight range.
2 S.D. would be the interval 7 to 15 ounces or about 95% (9500 boxes) of them.
3 S.D. would be in the interval 5 to 17 or about 99.7% (9970 boxes) of them.
So, we can see that 18 is 3.5 S.D. above the mean. Those are very heavy boxes of cereal, and the company does not many of those.
