A vending machine automatically pours soft drinks into cups. The amount of soft drink dispensed into a cup is normally distributed with a mean of 7.6oz and standard deviation of 0.4oz.

(a) Estimate the probability that the machine will overflow an 8-ounce cup. (Round your answer to two decimal places.)
(b) Estimate the probability that the machine will not overflow an 8-ounce cup. (Round your answer to two decimal places.)
(c) The machine has just been loaded with 848 cups. How many of these do you expect will overflow when served

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Answer:

Step-by-step explanation:

Since the amount of soft drink dispensed into a cup is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = amount in ounce of soft drink dispensed into cup.

µ = mean amount

σ = standard deviation

From the information given,

µ = 7.6oz

σ = 0.4 oz

a) The probability that the machine will overflow an 8-ounce cup is expressed as

P(x > 8) = 1 - P(x ≤ 8)

For x = 8,

z = (8 - 7.6)/0.4 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.84

P(x ≤ 8) = 1 - 0.84 = 0.16

b) P(x< 8) = 0.84

c) when the machine has just been loaded with 848 cups, the number of cups expected to overflow when served is

0.16 × 848 = 136 cups

Using the normal distribution, it is found that:

a) 0.1587 = 15.87% probability that the machine will overflow an 8-ounce cup.

b) 0.8413 = 84.13% probability that the machine will not overflow an 8-ounce cup.

c) 135 cups are expected to overflow when served.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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, which is the percentile of X.

In this problem:

  • Mean of 7.6 oz, thus [tex]\mu = 7.6[/tex].
  • Standard deviation of 0.4 oz, thus [tex]\sigma = 0.4[/tex].

Item a:

The probability of an overflow is the probability of the machine having 8 oz or more, thus, it is 1 subtracted by the p-value of Z when X = 8.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{8 - 7.6}{0.4}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a p-value of 0.8413.

1 - 0.8413 = 0.1587.

0.1587 = 15.87% probability that the machine will overflow an 8-ounce cup.

Item b:

0.1587 probability of an overflow, so 1 - 0.1587 = 0.8413 = 84.13% probability that the machine will not overflow an 8-ounce cup.

Item c:

0.1587 out of 848, thus:

[tex]0.1587(848) = 135[/tex]

135 cups are expected to overflow when served.

A similar problem is given at https://brainly.com/question/12490452