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Question 12
The annual rainful record of a city for 66 days is given in the following table.
Rainfall(in cm)0−1010−2020−3030−4040−5050−60Number of days221081556
Calculate the median rainfall using ogives (or move than type and of less than type)
The annual rainful record of a city for 66 days is given in the following table.
Rainfall(in cm)0−1010−2020−3030−4040−5050−60Number of days221081556
Calculate the median rainfall using ogives (or move than type and of less than type)
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Solution
We observe that the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10.
So, the total annual rainfall record of a city for less than 10cm is 0+22=22 days.
Continuing in this manner, we will get remaining less than 20,30,40,50, and 60.
Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. Since 22 days lies in the interval 0-10. So, annual rainfall record for 66-22=44 days is more than or equal to 10cm. Continuing in this manner we will get remaining more than or equal to 20, 30, 40, 50 and 60.
Now, we construct a table for less than and more than type.
Rainfall (in cm)Number of daysRainfall (in cm)Number of daysLess than 00More than or equal to 066Less than 100+22=22More than or equal to 1066−22=44Less than 2022+10=32More than or equal to 2044−10=34Less than 3032+8=40More than or equal to 3034−8=26Less than 4040+15=55More than or equal to 4026−15=11Less than 5055+5=60More than or equal to 5011−5=6Less than 6060+6=66More than or equal to 606−6=0
To draw less than type ogive we plot the points (0,0), (10,22), (20,32), (30,40), (40,55), (50,60), (60,66) on the paper and join them by free hand.
To draw the more than type ogive we plot the points (0,66), (10,44), (20,34), (30,26), (40,11), (50,6) and (60,0) on the graph paper and join them by free hand.
∵ Total number of days(n)=66
Now, n2=33
Firstly, we plot a line parallel to X-axis at an intersection point of both ogives, which further intersect at (0,33) on Y-axis. Now, we draw a line perpendicular to X-axis at an intersection point of both ogives, which further intersect at (21.25, 0) on X-axis, which is the required median using ogives.
Hence, median rainfall=21.25cm.
So, the total annual rainfall record of a city for less than 10cm is 0+22=22 days.
Continuing in this manner, we will get remaining less than 20,30,40,50, and 60.
Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. Since 22 days lies in the interval 0-10. So, annual rainfall record for 66-22=44 days is more than or equal to 10cm. Continuing in this manner we will get remaining more than or equal to 20, 30, 40, 50 and 60.
Now, we construct a table for less than and more than type.
Rainfall (in cm)Number of daysRainfall (in cm)Number of daysLess than 00More than or equal to 066Less than 100+22=22More than or equal to 1066−22=44Less than 2022+10=32More than or equal to 2044−10=34Less than 3032+8=40More than or equal to 3034−8=26Less than 4040+15=55More than or equal to 4026−15=11Less than 5055+5=60More than or equal to 5011−5=6Less than 6060+6=66More than or equal to 606−6=0
To draw less than type ogive we plot the points (0,0), (10,22), (20,32), (30,40), (40,55), (50,60), (60,66) on the paper and join them by free hand.
To draw the more than type ogive we plot the points (0,66), (10,44), (20,34), (30,26), (40,11), (50,6) and (60,0) on the graph paper and join them by free hand.
∵ Total number of days(n)=66
Now, n2=33
Firstly, we plot a line parallel to X-axis at an intersection point of both ogives, which further intersect at (0,33) on Y-axis. Now, we draw a line perpendicular to X-axis at an intersection point of both ogives, which further intersect at (21.25, 0) on X-axis, which is the required median using ogives.
Hence, median rainfall=21.25cm.
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