Answer:
a) 5.48% of the cups will contain more than 224 milliliters
b) 45.14% probability that a cup contains between 191 and 209 milliliters
c) 22.8(rounding up, 23) cups will probably overflow if 230- milliliter cups are used for the next 1000 drinks
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 200, \sigma = 15](https://tex.z-dn.net/?f=%5Cmu%20%3D%20200%2C%20%5Csigma%20%3D%2015)
(a) what fraction of the cups will contain more than 224 milliliters?
This is 1 subtracted by the pvalue of Z when X = 224. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{224 - 200}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B224%20-%20200%7D%7B15%7D)
![Z = 1.6](https://tex.z-dn.net/?f=Z%20%3D%201.6)
has a pvalue of 0.9452
1 - 0.9452 = 0.0548
5.48% of the cups will contain more than 224 milliliters
(b) what is the probability that a cup contains between 191 and 209 milliliters?
This is the pvalue of Z when X = 209 subtracted by the pvalue of Z when X = 191. So
X = 209
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{209 - 200}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B209%20-%20200%7D%7B15%7D)
![Z = 0.6](https://tex.z-dn.net/?f=Z%20%3D%200.6)
has a pvalue of 0.7257
X = 191
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{191 - 200}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B191%20-%20200%7D%7B15%7D)
![Z = -0.6](https://tex.z-dn.net/?f=Z%20%3D%20-0.6)
has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
45.14% probability that a cup contains between 191 and 209 milliliters
c) how many cups will probably overflow if 230- milliliter cups are used for the next 1000 drinks?
Proportion of cups with more than 230 milliliters.
This is 1 subtracted by the pvalue of Z when X = 230.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{230 - 200}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B230%20-%20200%7D%7B15%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
Out of 1000
0.0228*1000 = 22.8
22.8(rounding up, 23) cups will probably overflow if 230- milliliter cups are used for the next 1000 drinks