Question

Q. 1 In 1798 the scientist Henry Cavendish used a torsion balance to estimate the density of the earth as a multiple of the density of water. He ran his experiment 29 times and obtained the following results: 5.505.575.425.615.53 5.47 4.885.625.63 4.07 5.29 5.34 5.26 5.445.46 5.55 5.34 5.30 5.365.79 5.75 5.29 5.105.86 5.58 5.27 5.85 5.65 5.39 a) Construct an appropriate stem-and-leaf plot of this data b) Describe the distribution of the dataset based on the stem-and-leaf plot c) Calculate the mean and median of the dataset. Which do you think is the better measure of location for this dataset? Why? d) Calculate the 5% 10% and 20% trimmed means for this dataset. e) Calculate the inter-quartile range of the data.

Answer 1

Following table shows the calculations:

	X	Rounded
	4.07	4.1
	4.88	4.9
	5.1	5.1
	5.26	5.3
	5.27	5.3
	5.29	5.3
	5.29	5.3
	5.3	5.3
	5.34	5.3
	5.34	5.3
	5.36	5.4
	5.39	5.4
	5.42	5.4
	5.44	5.4
	5.46	5.5
	5.47	5.5
	5.5	5.5
	5.53	5.5
	5.55	5.6
	5.57	5.6
	5.58	5.6
	5.61	5.6
	5.62	5.6
	5.63	5.6
	5.65	5.7
	5.75	5.8
	5.79	5.8
	5.85	5.9
	5.86	5.9
Total	157.17

Following is the stem and leaf display:

b)

Stem and leaf plot shows that distribution is skewed to left and it is unimodal.

c)

The mean is:

Median is middle value of ordered data set. Since there are 29 data values so median will be 15th data value. That is median is 5.46.

Since distribution is not symmetric so median will be more appropriate.

d)

Since there are 29 data values so 29*0.05 = 1.45 or 2 data values from both end. Following is the trimmed data and sum of trimmed data:

	X
	5.1
	5.26
	5.27
	5.29
	5.29
	5.3
	5.34
	5.34
	5.36
	5.39
	5.42
	5.44
	5.46
	5.47
	5.5
	5.53
	5.55
	5.57
	5.58
	5.61
	5.62
	5.63
	5.65
	5.75
	5.79
Total	136.51

The 5% trimmed mean is :

----------------------------

Since there are 29 data values so 29*0.10 = 2.9 or 3 data values from both end. Following is the trimmed data and sum of trimmed data:

	X
	5.26
	5.27
	5.29
	5.29
	5.3
	5.34
	5.34
	5.36
	5.39
	5.42
	5.44
	5.46
	5.47
	5.5
	5.53
	5.55
	5.57
	5.58
	5.61
	5.62
	5.63
	5.65
	5.75
Total	125.62

The 10% trimmed mean is :

Since there are 29 data values so 29*0.20 = 5.8 or 6 data values from both end. Following is the trimmed data and sum of trimmed data:

	X
	5.29
	5.3
	5.34
	5.34
	5.36
	5.39
	5.42
	5.44
	5.46
	5.47
	5.5
	5.53
	5.55
	5.57
	5.58
	5.61
	5.62
Total	92.77

The 20% trimmed mean is :

e)

First quartile: First quartile is median of first half of data set. There are 15 data values (including median) in first half of data set. That is first quartile will 8th data values. Hence, first quartile is $Q_{1}=5.30$.

Third quartile: Third quartile is median of second half of data set. There are 15 data values (including median) in second half of data set. That is third quartile will be 22nd data values. Hence, third quartile is $Q_{3}=5.61$.

Intequartile range: