SOLUTION: The packaging process in a breakfast cereal company has been adjusted so that the average of m = 13.0 oz of cereal is placed in packages. Of course, not all packages have precisel
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Question 670683: The packaging process in a breakfast cereal company has been adjusted so that the average of m = 13.0 oz of cereal is placed in packages. Of course, not all packages have precisely 13.0 oz and the standard deviation of the actual weight is s = 0.1 oz and the distribution of weights is known to follow the normal probability distribution.
a) What proportion of the packages will contain more than 13.3 oz of cereal?
I know the following: Z = X-m/s = 13.3 - 13.0 / .1 = +3
So: P (x > 13.3) = P (Z > 3) = ? I don't know how to change the 3 into a proportion.
b) What proportion of the packages will contain less than 13.3 oz?
Z = X-m/s = 3. P (x < 13.3) = P(Z<3) = 1 - answer to a. Is this correct?
c. Calculate the 85th percentile and interpret the result. I know the answer is to the left of the 85th percentile but how do I find that?
d. Calculate the first quartile of this distribution. I don't know how to start.
Thanks! Answer by ewatrrr(24785) (Show Source):
Hi,
finding p-value after determining the correct z-value (or visa-versa)can be done:
I. using a chart
II. TI calculator
III. Excel function
a) z = 3 ⇒ p = .999 Or 99.9% ≤ 13.3oz
What proportion of the packages will contain more than 13.3 oz of cereal?
1- .999 = .001 Or 1%
Calculate the 85th percentile and interpret the result.
IF P = .85 ⇒ z = 1.04
1.04 = (x - 13)/.1
.104 + 13 = x
13.104 = x, the 85th percentile
Important to Understand z -values as they relate to the Standard Normal curve:
Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
Note: z = 0 , 50% of the area under the curve is to the left and 50% to the right
