Question

6.11 A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, (a) what fraction of the cups will contain more than 224 milliliters? (b) what is the 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? (d) below wh

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 200, \sigma = 15$

(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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{224 - 200}{15}$

$Z = 1.6$

$Z = 1.6$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{209 - 200}{15}$

$Z = 0.6$

$Z = 0.6$ has a pvalue of 0.7257

X = 191

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{191 - 200}{15}$

$Z = -0.6$

$Z = -0.6$ 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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{230 - 200}{15}$

$Z = 2$

$Z = 2$ 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