Question

The enrollments of all 13 public universities in the state of ohio are listed below. college enrollment university of akron 25,612 bowling green state university 18,641 central state university 1,725 university of cincinnati 36,502 cleveland state university 15,693 kent state university 34,187 miami university 17,338 ohio state university 59,501 ohio university 20,151 shawnee state university 4,344 university of toledo 20,839 wright state university 18,670 youngstown state university 14,655 (in parts b-e, do not round any intermediate calculations. round your final answers to the nearest whole number). (a) is this a sample or a population? (b) what is the mean enrollment? mean 22143 (c) what is the median enrollment? median 18670 (d) what is the range of the enrollments? range 57776 (e) compute the standard deviation. standard deviation rev: 01_03_2013, 01_17_2013 ©2018 mcgraw-hill education. all rights reserved.

Answer 1

Answers:

(a) The enrollment data is a population
(b) Mean: 22,150.62
(c) Median: 18,670
(d) Range: 57,776
(e) Standard Deviation: 14,258.49

Explanations:

(a) As we noticed in the problem, the enrollment data is used for the study of the number of enrollments in all of the public universities in Ohio. Since the given enrollment data contains the number of enrollments in all of the public univeristies in Ohio, the enrollment data is a population.

(b) To compute the mean, we get the sum of all the data points and divide it by the number of data points.

Since,

Sum of all data points =  1725 + 4344 + 14655 + 15693 + 17338 + 18641 18670 + 20251 + 20839 + 25612 + 34187 + 36502  +  59,501
Sum of all data points =  287,958

and the number of data points = 13,

mean = 287,958 ÷ 13 =  22,150.62

(c) The median is computed when the data points are arranged from least to greatest and it is equal to either:

>> the middle number - if the number of data points is an odd number, or
>> the average of 2 middle numbers - if the number of data points is an even number

In the enrollment data, when the data points are arranged from lowest to highest, the following is the result:

1725, 4344, 14655, 15693, 17338, 18641, 18670, 20251, 20839, 25612, 34187, 36502, 59,501

Since the number of data points is 13 and 13 is an odd number, we find the middle number, which is the 7th lowest number = 18670.

Hence, the median is 18670.

(d) The range of the data is the difference between the highest data point and the lowest data point. 

Since the highest data point is 59,501, the lowest data point is 1725 and their difference is 59,501 - 1725 =  57,776. The range of the enrollment data is  57,776.

(e) To compute for standard deviation,

>> First, we compute the squares of all the data points and find their sum:

Sum of the squares of data points =  1725² + 4344² + 14655² + 15693² + 17338² + 18641² + 18670² + 20251² + 20839² + 25612² + 34187² + 36502²  +  59,501² 

Sum of the squares of data points =  9,021,404,700 

>> Then, we compute the average of the squares of all data points by dividing the sum of the squares of all data points by the number of data points:

Average of the squares
= Sum of the squares of data points ÷ number of data points
= 9,021,404,700 ÷ 13
Average of the squares = 693,954,207.69231

>> Finally, the standard deviation is the square root of the difference of the average of squares and the square of the average of the data points:

$\sigma = \sqrt{E(X^2) -(E(X))^2}$

where:

$\sigma = \text{standard deviation} \\ E(X) = \text{average of all data points} = 22,150.62 \\ E(X^2) = \text{average of the squares of all data points} \approx 693,954,207.69231$

So, the standard deviation is given by 

$\sigma = \sqrt{E(X^2) -(E(X))^2} \\ \sigma = \sqrt{693,954,207.69231 -(22,150.62)^2} \\ \boxed{\sigma \approx 14,258.49}$

Therefore, the standard deviation is 14,258.49.