Q-7 The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Run scored by two teams A and BRun scored by two teams A and B ball-wise in a 60 ball cricket match
Number of balls
Teams A
Team B
1 – 6
2
5
7 – 12
1
6
13 – 18
8
2
19 – 24
9
10
25 – 30
4
5
31 – 36
5
6
37 – 42
6
3
43 – 48
10
4
49 – 54
6
8
55 – 60
2
10
Represent the table data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Solution:
It can be observed that the class intervals of the given data are not continuous.
There is a gap of 1 ball between them. Therefore, we elongate each interval by 1 ball by subtracting ½ from lower limit and adding ½ to upper class intervals.
0.5 has to be subtracted from the lower class limits.
Let’s calculate the class marks as. Following table shows the class marks.
Runs scored in range of balls with class marksClass marks calculated for last table
Number of balls
Class marks
Team A
Team B
0.5-6.5
3.5
2
5
6.5-12.5
9.5
1
6
12.5-18.5
15.5
8
2
18.5-24.5
21.5
9
10
24.5-30.5
27.5
4
5
30.5-36.5
27.5
4
5
36.5-42.5
39.5
6
3
42.5-48.5
45.5
10
4
48.5-54.5
51.5
6
8
54.5-60.5
57.5
2
10
Plot the histogram,
Plot the class marks on x-axis class mark (balls) and runs scored on y-axis.
Again choose the scale so that 1 run is equal to 1 unit on y-axis.
Draw the following graph
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Q-8 A random survey of the number of children of various age groups playing in park was found as follows:
Frequency distribution of ages of childrenFrequency distribution of ages of children found using random survey of children playing in park
Age (in years)
Number of children
1-2
5
2-3
3
3-5
6
5-7
12
7-10
9
10-15
10
15-17
4
Draw a histogram to represent the data above.
Solution: The problem with this data is that the class intervals are of varying width. The principle behind histograms is that the area of each bar represents the fraction of a frequency (probability) distribution within each bin (class, interval). Therefore, as the bar in a histogram is spread across bins, it is squished in height (that is, area remains the same). Here we choose the bin size to be 1 year. So we divide the frequency (number of children) into bins of 1 year.
For example, number of children in range 1-2 is 5 however since 2-1 = 1 this 5 is divided into one bin of 1 year, i.e. it is represented in our histogram unchanged. However for the range 3 – 5 the difference is 2 years 5 – 3 = 2. Therefore the frequency (number of children) of 6, has to be divided into two bins of 1 year each. That is, we divide this 6 in two portions, height of each of these portions would be . The formula in general is Height of histogram rectangle =
Heights of rectanges calulated in inverse proportion to unit intervalsHeights of rectanges calculated in inverse proportion to unit intervals
Age (In year)
Frequency (number of children)
Width of classes
Height of rectangle
1-2
5
1
2-3
3
1
3-5
6
2
5-7
12
2
7-10
9
3
10-15
10
5
15-17
4
2
Let’s plot the chart below, with following details:
Put the intervals of ages of children on x-axis.
Plot the height of histogram from the above table.
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