The fill volume of an automated filling machine used for filling cans of carbona

Question

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces and a standard deviation of 0.1 fluid ounce. Use the 68-95-99.7% Rule, not the Z table, to answer the following questions. a) What is the probability that a fill volume is less than 12.1 fluid ounces? b) What is the probability that a fill volume is between 12.2 and 12.6 fluid ounces? c) What is the probability that a fill volume is greater than 12.5 fluid ounces? d) A random sample of 100 cans was selected. What is the probability that the sample mean fill volume is greater than 12.42 fluid ounces? A random sample of 100 cans was selected. What is the probability that the sample mean fill volume is between 12.39 and 12.41 fluid ounces? e)

Explanation / Answer

Ans:

Given that

mean=12.4

standard deviation=0.1

a)12.4-12.1=0.3

so,12.1 is 3 standard deviation below the means.

As,99.7% of the data lies within 3 standard deviation of the mean,half of the rest of the data i.e. 0.5*(1-0.997)=0.0015 or 0.15% lie below 12.1

0.15%

b)12.2 and 12.6 is 2 standard deviations below and above the mean.

95% of the data lies within 2 standard deviations of the mean i.e. between 12.2 and 12.4

95%

c)12.5 is one standard deviation of the mean.

As,68% data lies within one standard deviation of the mean,so half of the rest of the data i.e. 0.5*(1-0.68)=0.16 or 16% will lie above 12.5

16%

d)standard error of mean=0.1/sqrt(100)=0.01

12.42 is 2 standard deviation above the mean.

As,95% data lies within 2 standard deviation of the mean,so half of the rest of the data i.e. 0.5*(1-0.95)=0.025 or 2.5% will lie above 12.42

2.5%

e)standard error of mean=0.1/sqrt(100)=0.01

12.39 and 12.41 are one standard deviation below and above the mean,so 68% of the data will lie between 12.39 and 12.41.

68%

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