Question

For the follwoing date find:

30	20	35	17	50	54
64	29	34	23	20	20
21	50	26	26	20	48
20	30	23	29	17	17
20	18	23	36	20	20

1. The Mean=
2. The Median=
3. The Mode=
4. The Range =
5. The Standard Deviation is =
6. The Variance is=
7. Q1Q1 =
8. Q3Q3 =
9. IQR=
10. Upper Fence=
11. Lower Fence is =
12.If you graphed the data above, what shape would you expect the data to be?

• Skewed-Right
• Bell-Shaped
• Skewed-Left

Answer 1

Solution:

For the given data, we have to find measures of central tendency and variation. The calculation table required for the calculation purpose is given as below:

No.	X	(X - mean)^2
1	17	136.1111889
2	17	136.1111889
3	17	136.1111889
4	18	113.7778489
5	20	75.11116889
6	20	75.11116889
7	20	75.11116889
8	20	75.11116889
9	20	75.11116889
10	20	75.11116889
11	20	75.11116889
12	20	75.11116889
13	21	58.77782889
14	23	32.11114889
15	23	32.11114889
16	23	32.11114889
17	26	7.111128889
18	26	7.111128889
19	29	0.111108889
20	29	0.111108889
21	30	1.777768889
22	30	1.777768889
23	34	28.44440889
24	35	40.11106889
25	36	53.77772889
26	48	373.7776489
27	50	455.1109689
28	50	455.1109689
29	54	641.7776089
30	64	1248.444209
Total	860	4592.666667
Mean	28.66667

We have

Sample size = n = 30

Mean = ∑X/n = 860/30 = 28.66667

Median = Middle most observation when data is in an increasing or decreasing order

Here, n = 30, so median is the mean of 15^th and 16^th observations.

Median = (23 + 23)/2 = 23

Median = 23

Mode = Observation with highest frequency

Mode = 20

Minimum = 17

Maximum = 64

Range = Maximum – Minimum = 64 – 17 = 47

Standard deviation = sqrt[∑(X – mean)^2/(n – 1)]

Standard deviation = sqrt[4592.666667/(30 – 1)]

Standard deviation = sqrt(158.3678161)

Standard deviation = 12.58442752

Variance = ∑(X – mean)^2/(n – 1)

Variance = 4592.666667/(30 – 1)

Variance = 158.3678161

Q1 = 20

Q3 = 34

(by using excel)

IQR = Q3 – Q1 = 34 – 20 = 14

Upper Fence = Q3 + 1.5*IQR = 34 + 1.5*14 = 55

Lower Fence = Q1 – 1.5*IQR = 20 - 1.5*14 = -1

Boxplot for the given data is given as below:

From above box plot, it is observed that the given data is skewed at right.