Question

The following distribution gives the weights of 60 students in a class. Find the mean and mode weights of the students.

Weight (in kg)	40-44	44-48	48-52	52-56	56-60	60-64	64-68	68-72
Number of students	4	6	10	14	10	8	6	2

Answer 1

Hint: First of all, calculate the class mark for each class interval using the formula, \[\text{Class}\,\text{Mark=}\dfrac{\text{actual}\,\text{upper}\,\text{limit+actual lower}\,\text{limit}}{\text{2}}\] . Here, frequency \[\left( {{f}_{i}} \right)\] is the row of “Number of students' '. Now, modify the given table by adding two more columns that are class marked \[\left( {{x}_{i}} \right)\] and \[{{f}_{i}}{{x}_{i}}\] . Use the formula, \[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] to calculate the mean. Then, use the formula, \[Mode=l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\] to calculate the mode. Solve it further and get the required answer.

Complete step-by-step solution:
According to the question, we have a distribution table that shows the weights of 60 students.
Here, frequency \[\left( {{f}_{i}} \right)\] is the row of “Number of students”.
We know that the mid value of every class interval is called its class mark and it can be calculated by using the formula, \[\text{Class}\,\text{Mark=}\dfrac{\text{actual}\,\text{upper}\,\text{limit+actual lower}\,\text{limit}}{\text{2}}\] ………………………………………………(1)
Now, using the formula shown in equation (1), we get
The class mark for the class interval, 40-44 = 42
The class mark for the class interval, 44-48 = 46
The class mark for the class interval, 48-52 = 50
The class mark for the class interval, 52-56= 54
The class mark for the class interval, 56-60 = 58
The class mark for the class interval, 60-64 = 62
The class mark for the class interval, 64-68 = 66
Now, using the above data and modifying the given table, we get

Weight (in kg)	Number of students \[\left( {{f}_{i}} \right)\]	Class mark \[\left( {{x}_{i}} \right)\]	\[{{f}_{i}}{{x}_{i}}\]
40-44	4	42	168
44-48	6	46	276
48-52	10	50	500
52-56	14	54	756
56-60	10	58	580
60-64	8	62	496
64-68	6	66	396
68-72	2	70	140

We know the formula for the mean, \[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] ……………………………………………(2)
Now, on using the data from the modified table and the formula shown in equation (2), we get
\[Mean=\dfrac{168+276+500+756+580+496+396+140}{4+6+10+14+10+8+6+2}=\dfrac{3312}{60}=55.2\] ……………………………………….(3)
We also know the formula, \[Mode=l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\] …………………………………………(4)
where,
l = lower limit of the modal class,
h = size of the class interval,
\[{{f}_{1}}\] = frequency of the modal class,
\[{{f}_{0}}\] = frequency of the class preceding the modal class,
\[{{f}_{2}}\] = frequency of the class succeeding the modal class.
Since the class interval 52-56 has the highest frequency so, the modal class of the given distribution is 52-56.
Now, from the above table,
The lower limit of the modal class, \[l\] = 52 …………………………………………(5)
 The size of the class interval of modal class, h = \[56-52=4\] ………………………………………………(6)
The frequency of the modal class, \[{{f}_{1}}\] =14 …………………………………………(7)
The frequency of the class preceding the modal class, \[{{f}_{0}}\] = 10 ……………………………………………(8)
 The frequency of the class succeeding the modal class, \[{{f}_{2}}\] = 10 ………………………………………………(9)
Now, from equation (4), equation (5), equation (6), equation (7), equation (8), and equation (9), we get
\[Mode=52+\left( \dfrac{14-10}{2\times 14-10-10} \right)\times 4=52+\left( \dfrac{4}{8} \right)\times 4=54\] ……………………………………………….(10)
From equation (3) and equation (10), we have mean and mode.
Therefore, the mean and mode of the given distribution are 55.2 and 54 respectively.

Note: It is very complex to solve this question without using the formula. Therefore, to solve this question, always keep in mind the formula for the mean and median that are \[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] and \[Mode=l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\] respectively.