Question 2
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)Number of studentsLess than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235
Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph verify the result by using the formula.
Question 2
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)Number of studentsLess than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235
Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph verify the result by using the formula.
The given cumulative frequency distributions of less than type is -
Weight (in kg) −upper class limitsNumber of students (cumulative frequency)Less than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235
Now, taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following -
Total number of students=n=35
n/2 = 17.5
Now, mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So, the median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies as below.
Weight (in kg)Frequency (f)Cumulative frequencyLess than 380038−403−0=3340−425−3=2542−449−5=4944−4614−9=51446−4828−14=142848−5032−28=43250−5235−32=335Total (n)35
Now, the cumulative frequency just greater than n2(i.e,352=17.5) is 28 belonging to class interval 46 - 48.
Median class = 46 - 48
Lower class limit l of median class = 46
Frequency f of median class = 14
Cumulative frequency cf of class preceding median class = 14
Class size h = 2
Median=l+(n2−cff)×h
=46+(17.5−1414)×2
=46+3.57
=46.5
So, median of this data is 46.5.
Hence, value of median is verified.
The given cumulative frequency distributions of less than type is -
Weight (in kg) −upper class limitsNumber of students (cumulative frequency)Less than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235
Now, taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following -
Total number of students=n=35
n/2 = 17.5
Now, mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So, the median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies as below.
Weight (in kg)Frequency (f)Cumulative frequencyLess than 380038−403−0=3340−425−3=2542−449−5=4944−4614−9=51446−4828−14=142848−5032−28=43250−5235−32=335Total (n)35
Now, the cumulative frequency just greater than n2(i.e,352=17.5) is 28 belonging to class interval 46 - 48.
Median class = 46 - 48
Lower class limit l of median class = 46
Frequency f of median class = 14
Cumulative frequency cf of class preceding median class = 14
Class size h = 2
Median=l+(n2−cff)×h
=46+(17.5−1414)×2
=46+3.57
=46.5
So, median of this data is 46.5.
Hence, value of median is verified.
