110 98 119 105 99 100 100 110
Compute the ff. using ungrouped data
V. The following data represent the respective heights (in inches) of 30 students enrolled in a physical education class. Construct a frequency distribution with eight class intervals.
60.0 59.7 70.1 58.2 62.1
60.1 62.3 62.4 68.3 63.3
67.3 62.7 58.0 67.3 72.8
64.4 67.1 66.4 61.7 69.4
65.3 65.7 66.3 68.1 68.3
65.4 66.7 61.4 64.1 69.2
Person |
Height |
Self Esteem |
1 |
68 |
4.1 |
2 |
71 |
4.6 |
3 |
62 |
3.8 |
4 |
75 |
4.4 |
5 |
58 |
3.2 |
6 |
60 |
3.1 |
7 |
67 |
3.8 |
8 |
68 |
4.1 |
9 |
71 |
4.3 |
10 |
69 |
3.7 |
11 |
68 |
3.5 |
12 |
67 |
3.2 |
13 |
63 |
3.7 |
14 |
62 |
3.3 |
15 |
60 |
3.4 |
16 |
63 |
4.0 |
17 |
65 |
4.1 |
18 |
67 |
3.8 |
19 |
63 |
3.4 |
20 |
61 |
3.6 |
1. Here we have to compute s
ummary statistics i.e. mean, median, mode, variance, standard deviation.
For mean we have, where Xi is observations and N is total no of observations.
So we have, = (110+98+119+105+99+100+100+110)/8 =92.625=mean.
For mode, we know that it is the value that occurs max no of time in data, hence mode is 100
Median is the value which occurs in the middle of the list of values given hence, on rearranging the values we get, 98,99,100,100,105,110,110,119. We can see that 100 is the middle most value hence it is the median of given data.
Given data: Weight of female nurses in a hospital in lbs. 110 98 119 105 99 &nbs
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