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Gekata [30.6K]
2 years ago
7

A cola-dispensing machine is set to dispense a mean of 2.02 liters into a bottle labeled 2 liters. Actual quantities dispensed v

ary and the amounts are normally distributed with a standard deviation of 0.015 liters. a. What is the probability a bottle will contain between 2.00 and 2.03 liters? b. What is the probability a bottle will contain less than 2 liters? c. 2% of the containers will contain how much cola or more?
Mathematics
1 answer:
sammy [17]2 years ago
5 0

Answer:

Step-by-step explanation:

Given that a  cola-dispensing machine is set to dispense a mean of 2.02 liters into a bottle labeled 2 liters.

Std deviation =0.015 litres

X- litres contained in a bottle is N(2.02, 0.15)

Z score is obtained as z=\frac{x-2.02}{0.015}

a) probability a bottle will contain between 2.00 and 2.03 liters

=P(2<x<2.03) = P(-1.33<Z<2)

= 0.4082+0.4772

=0.8854

b) P(X<2) = P(Z<-1.33) =0.5-0.4082 = 0.0918

c) 2% of containers

|z|<0.11

X lies between 0.6883 and 3.352 l

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we know that

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In a production process, the diameter measures of manufactured o-ring gaskets are known to be normally distributed with a mean d
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Answer:

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OTHER WAY

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And we can find this probability using the normal standard table or excel:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the diameter of a population, and for this case we know the distribution for X is given by:

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And we know that if the diameter is 75 or less the ring would be considered defective , so then in order to find the proportion of defective we need to find the following probability:

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One way to do this in excel is with the following formula:

"=NORM.DIST(75,80,3,TRUE)"

And we got: P(X\leq 75)= 0.04779

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And we can find this probability using the normal standard table or excel:

P(Z

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