Following are the ages of $360$ patients getting medical treatment in a hospital on a day:

Age (in years)	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
No. of Patients	$90$	$50$	$60$	$80$	$50$	$30$

Construct a cumulative frequency distribution.

Hint: In this problem we need to construct the cumulative frequency distribution for the given data. For the cumulative distribution table we need to consider the upper boundaries of each class. The cumulative frequency of each class can be calculated by adding the cumulative frequency of preceding class and the frequency of the current class. The cumulative frequency of the first class is always equal to the frequency of the class. Likewise, we will calculate all the cumulative frequencies of the classes and write them in a table form.

Complete step by step answer:
Given data,


Consider the class $10-20$, the upper boundary of the class is $20$. Frequency of the class is $90$.
We can observe that the class $10-20$ is the first class in the given data, so the cumulative frequency of the class $10-20$ is $90$.
Consider the class $20-30$, the upper boundary of the class is $30$. Frequency of the class is $50$.
The preceding class of the class $20-30$ is $10-20$. The cumulative frequency of the preceding the class $10-20$ is $90$.
So, the cumulative frequency of the class $20-30$ is given by $50+90=140$.
Consider the class $30-40$, the upper boundary of the class is $40$. Frequency of the class is $60$.
The preceding class of the class $30-40$ is $20-30$. The cumulative frequency of the preceding the class $20-30$ is $140$.
So, the cumulative frequency of the class $30-40$ is given by $140+60=200$.
Consider the class $40-50$, the upper boundary of the class is $50$. Frequency of the class is $80$.
The preceding class of the class $40-50$ is $30-40$. The cumulative frequency of the preceding the class $30-40$ is $200$.
So, the cumulative frequency of the class $40-50$ is given by $200+80=280$.
Consider the class $50-60$, the upper boundary of the class is $60$. Frequency of the class is $50$.
The preceding class of the class $50-60$ is $40-50$. The cumulative frequency of the preceding the class $40-50$ is $280$.
So, the cumulative frequency of the class $50-60$ is given by $280+50=330$.
Consider the class $60-70$, the upper boundary of the class is $70$. Frequency of the class is $30$.
The preceding class of the class $60-70$ is $50-60$. The cumulative frequency of the preceding the class $50-60$ is $330$.
So, the cumulative frequency of the class $60-70$ is given by $330+30=360$.
Now the cumulative frequency distribution table for the given data is


Note: After constructing a cumulative frequency table we can also check whether the obtained table is correct or not. When we add all the frequencies of the classes that value should be equal to the cumulative frequency of the last class, then only our distribution table is correct.