Question

In: Statistics and Probability

Stem-and-leaf of Tuition costs (thousands of dol N = 40 1 1 5 1 1 5...

Stem-and-leaf of Tuition costs (thousands of dol N = 40

1	1	5
1	1
5	1	8999
12	2	0000111
(10)	2	2222333333
18	2	4445555
11	2	66
9	2	88899
4	3	000
1	3	2

Leaf Unit = 1

a. Did any colleges have a tuition of $20,000? If so, how many?

b. How many colleges have a tuition rate greater than $28,000?

c. Would you say that this data set has an outlier? Why or why not?

d. Based on the answer above, which of the following do you think would be true for this data set AND why? (Mean = Median, Mean<Median, Mean>Median)

e. Based on your answers to the above questions do you believe this data set is normal, skewed right or skewed left?

Expert Solution

a) Yes, some colleges have a tuition cost of $20000. There are 4 colleges with a tuition of $20000.

b) There are 6 colleges with tuition rat greater than $28000.

c) The median of the tuition cost is = $23000

The first quartile = $21000

The third quartile = $26000

IQR = Q3 - Q1 = $5000

The upper limit for outliers = $26000 + 1.5 * $5000 = $33500

The lower limit for the outliers = $21000 - 1.5 * $5000 = $13500

Since no value of the tuition cost is outside the limits, hence there are no outliers.

d) Based on the above result we can say that Mean = Median.

Explanation: Since there are no outliers hence we can be sure that there are no unusually high value or unusually low value. If there was an unusually low value then the mean would have been more inclined towards the lower value, but since the median is not influenced by outliers hence Mean < Median in this case. If there was an unusually high value then the mean would have been more inclined towards the higher value, but since the median is not influenced by outliers hence Mean > Median in this case. But in this question, there are no outliers hence we can say that Mean = Median.

e) Since there are no outliers hence there should be no significant skewness in the data set, neither right or left. And since we have established that Mean = Median, hence, based on the answers to the above question, we can say that the data set is normal.

Thank You!! Please Upvote!!

orchestra answered 1 year ago

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