Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 200, \sigma = 15[/tex]
(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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{224 - 200}{15}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{209 - 200}{15}[/tex]
[tex]Z = 0.6[/tex]
[tex]Z = 0.6[/tex] has a pvalue of 0.7257
X = 191
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{191 - 200}{15}[/tex]
[tex]Z = -0.6[/tex]
[tex]Z = -0.6[/tex] 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{230 - 200}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] 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
