A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products. They hope to design future quality control tests to satisfy several criteria:  The calculated error for a sample set (the error is the difference between the true mean and mean of the sample set) does not exceed 2.0 g.  To save money on those tests, they want to collect as few samples as possible.  They want more than 90% of the confidence intervals they calculate to include the true mean. What value of n would you recommend for future tests?

So the answer for this case would be n=5 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

represent the sample mean

population mean (variable of interest)

s=2.5 represent the sample standard deviation

n represent the sample size  

ME = 2 represent the margin of error accepted

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

The margin of error is given by this formula:

   (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

  (3)

We can use as estimator for the population deviation the sample deviation

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got , replacing into formula (3) we got:

So the answer for this case would be n=5 rounded up to the nearest integer

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

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