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Anna71 [15]
4 years ago
10

The Walden University has 47,456 students in 2010, while Kaplan university had 77,966 students. Complete the following:

Mathematics
1 answer:
Umnica [9.8K]4 years ago
7 0

We have been given that The Walden University had 47,456 students in 2010, while Kaplan university had 77,966 students. We are required to compare the number of students in each of these universities in different ways.

(A) Let us first find difference between the number of students between the two universities. The difference between the number of students for two universities is 30510.

Let us now find what percentage is 30510 of 47456.

\frac{30510}{47456}\times 100 = 64.29 \%

Therefore, Kaplan’s enrollment was 64.29% larger than Walden’s

(B) In order to answer this part, we need to find what percentage is 30510 of 77966.

\frac{30510}{77966}\times 100 = 39.13 \%

Hence, Walden’s enrollment was 39.13% smaller than Kaplan’s.

(C) In this part we need to figure out what percentage is 47456 of 77966.

\frac{47456}{77966}\times 100 = 60.87 \%

Thus, Walden’s enrollment was 60.87% of Kaplan’s.

Note: Please round your answers to required number of decimal values.

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Answer:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=12 represent the population standard deviation

n represent the sample size  

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

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n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

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