Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Use graph paper for this question. A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded.Height (in cm)135-140140-145145-150150-155155-160160-165165-170No. of boys482014761Taking 2cm = height of 10cm along one axis and 2cm = 10 boys along the other axis. Draw an ogive of the distribution. Use the graph to estimate the median.

Last updated date: 24th Jun 2024
Total views: 375k
Views today: 6.75k
Verified
375k+ views
Hint: To, draw an ogive of the given data set, we will have to find the cumulative frequency at each interval. After that, we can plot a graph of marks vs cumulative frequency. Here, as it is not mentioned whether to draw a “less than” ogive or a “more than” ogive, we can draw a simple “less than” ogive where the cumulative frequency increases with each class considered.
Then to find the median we will just take half of the total frequency and mark that point image in the graphs.

Complete Step by Step Solution:
As we will have to find out the cumulative frequency at each class and then plot a graph using the values of cumulative frequency and marks, it is best to draw a table with three columns as class interval, frequency, and cumulative frequency.
 Height (in cm) Frequency Cumulative Frequency 135 – 140 4 4 140 – 145 8 12 145 – 150 20 32 150 – 155 14 46 155 – 160 7 53 160 – 165 6 59 165 – 170 1 60

Now, as we have the values of cumulative frequency at each interval, we can plot a graph between the upper limit of each interval and the cumulative frequency at that class. Remember, the reason why we are plotting the graph with the upper limit of the interval is that we are drawing a “less than” type ogive where the cumulative frequency at an interval is less than the upper limit of that interval.

Now, for the median, we will have to find $\dfrac{N}{2}$, where N is the total frequency which is equal to 60 here.
Now, we will draw a straight line from 30, and wherever it will meet on the x-axis, that will be our median.

Hence, the median is 29.9.

Note: We should plot the upper-class limits, not the lower-class limits. If we plot lower class limits it would give the wrong ogive of the distribution. We should plot cumulative frequency not frequency. Do not plot cumulative frequency on X-axis and the upper limit on Y-axis.