Construct a histogram from the following distribution of total marks obtained by 65 students of X class in the final examination
Marks(mid−points):150160170180190200No. of students810251273
Ascertainment of lower and upper class limits:
Since the difference between the second and first mid – point is 160 -150 =10
∴h=10⇒h2=5
So, lower and upper limits of the first class are 150 -5 and 150 + 5 i.e. 145 and 155 respectively
∴ First class interval is 145 – 155
Using the same procedure, we get the classes of other mid-points as under:
Marks:145−155155−165165−175175−185185−195195−205No. of students810251273
The histogram of the above frequency distribution is given in the figure.