Question

Given the following data, complete the following

64           47           33           61           43           13           74           50           41           42

66           65           54           61           49           1              25           32           61          

1. Create leaf and stem
2. (3pts) Create a frequency distribution with 4 classes (label everything)
3. Range
4. Find the mode
5. Find the mean
6. Find the variance
7. Find the standard deviation
8. (5pts) Find Tukey’s five number summery
9. Create a box and whisker plot
10. Find the IQR
11. Prove there are any outliers or not

Answer 1

Number of observations, n = 19

The Stem – Leaf plot is given below –

Stem	Leaf
0	1
1	3
2	5
3	2, 3
4	1, 2, 3, 7, 9
5	0, 4
6	1, 1, 1, 4, 5, 6
7	4

The frequency distribution is given below –

Classes	Frequency
1-20	2
21-40	3
41-60	7
61-80	7
Total	19

The Range of the data set = Maximum value – Minimum value = 74 – 1 = 73

Mode is that value of the data set that is repeated the maximum number of times.

Mode = 61 (repeated thrice)

The Excel Output of the following data set is given below -

А B с D E F G H 1 1 64 2 47 3 33 Total 882 Mean 46.42105 Variance 347.4017 Standard Deviation 4 5 18.63871 6 7 8 9 61 43 13 74 50 41 42 66 65 54 61 10 11 12 13 14 15 16 17 49 1 25 18 19 20 32 61 21

Therefore,

Mean = 46.421

Variance = 347.402

Standard Deviation = 18.6387

and the formulas used are -

A B с D E F G 1 64 47 2 3 33 Total Mean Variance Standard Deviation =SUM(B1:B19) =AVERAGE(B1:B19) =VAR.P(B1:B19) 4 61 5 ESTDEV.P(B1:B19) 6 43 13 74 50 7 8 9 10 41 42 11 66 65 12 13 54 14 61 49 15 16 1 17 25 32 18 19 61 20 21 22 Sheet1

where B1 and B19 are the cell numbers

Therefore,

Tukey's Five Number Summary is -

Maximum Value = 74

Minimum Value = 1

Median of the data set = 10th Observation

From the Leaf – Stem plot the 10th Observation is 49

First/Lower quartile of the data set is the median of the first half of the data set (that is the first 9 observations)

First Quartile = 5th Observation = 33

Third/Upper quartile of the data set is the median of the next half of the data set (that is from 11th observation to last observation)

Third Quartile = 5th Observation

Taking 11th Observation as the first Observation and counting from there-on,

5th Observation = 61

The Box - Whisker plot is given below -

乐 Trof (3)

Inter Quartile Range (IQR) = Third Quartile – First Quartile = 61 – 33 = 28

To calculate for outliers,

We multiply 1.5 to the IQR. Let p = 1.5 x IQR. Then we add the value p to the Third Quartile and also subtract p from the First Quartile to get a range.

If all the values of the data set lie inside this range then there are no outliers in the data set.

Therefore, p = 1.5 x 28 = 42

(p + Third Quartile) = 42 + 61 = 103

(Lower Quartile - p) = 33 – 42 = -9

The range of the data set is [-9, 103].

Since, all the values of the data set lie in this range there are no outliers in the data set.

(Since, nothing is mentioned the mean, variance and standard deviation are calculated in excel)