
Table below shows the number of students in the Maths Club of school classified according to their height.
HEIGHT (in cm) NUMBER OF STUDENTS 110-120 4 120-130 24 130-140 20 140-150 32 150-160 20
Calculate the mean height.
| HEIGHT (in cm) | NUMBER OF STUDENTS |
| 110-120 | 4 |
| 120-130 | 24 |
| 130-140 | 20 |
| 140-150 | 32 |
| 150-160 | 20 |
Answer
562.8k+ views
Hint:
We are given a class of heights of students of the Maths Club of school. And also the number of students belonging or falling in respective classes. We will find the mid-value(x) of the interval . Since here values of the midpoint of class are too big to calculate we will use the assumed mean method. In this method we will consider an assumed mean (a) from the values given in the first column of class. And also a column of deviation (d) will be added here. So let’s start!
Complete step by step solution:
We will consider the assumed mean to be a=135. There is no specific reason for it. We have to select a value such that it makes our calculations easy!
Now tabulate the data.
We have tabulated all values well in our data table above. Now we will use the formula to calculate the mean by the method assumed mean.
\[mean = a + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}\]
Now let’s find the terms of the formula.
\[\sum {{f_i}{d_i}} = \left( { - 80} \right) + \left( { - 240} \right) + 0 + 320 + 400 = 400\]
\[\sum {{f_i} = 4 + 24 + 20 + 32 + 20 = 100} \]
Now putting the values in formula we get,
\[
mean = a + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \\
\Rightarrow 135 + \dfrac{{400}}{{100}} \\
\Rightarrow 135 + 4 \\
\Rightarrow 139 \\
\]
Thus the mean height of students is \[ \Rightarrow 139cm\].
Note:
We can go with our traditional formula of finding the mean but that will make our work too tedious so used this assumed mean method. Also note that when you find the term \[{f_i}{d_i}\] we should be aware of negative signs also. Because that will affect when we find the summation.
We are given a class of heights of students of the Maths Club of school. And also the number of students belonging or falling in respective classes. We will find the mid-value(x) of the interval . Since here values of the midpoint of class are too big to calculate we will use the assumed mean method. In this method we will consider an assumed mean (a) from the values given in the first column of class. And also a column of deviation (d) will be added here. So let’s start!
Complete step by step solution:
We will consider the assumed mean to be a=135. There is no specific reason for it. We have to select a value such that it makes our calculations easy!
Now tabulate the data.
| HEIGHT(cm) | Number of students(f) | Mid-value(x) | \[{d_i} = x - a\] | \[{f_i}{d_i}\] |
| 110-120 | 4 | 115 | 115-135=-20 | \[4 \times \left( { - 20} \right) = - 80\] |
| 120-130 | 24 | 125 | 125-135=-10 | \[24 \times \left( { - 10} \right) = - 240\] |
| 130-140 | 20 | 135 | 135-135=0 | \[0\] |
| 140-150 | 32 | 145 | 145-135=10 | \[32 \times 10 = 320\] |
| 150-160 | 20 | 155 | 155-135=20 | \[20 \times 20 = 400\] |
| TOTAL | 100 | 400 |
We have tabulated all values well in our data table above. Now we will use the formula to calculate the mean by the method assumed mean.
\[mean = a + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}\]
Now let’s find the terms of the formula.
\[\sum {{f_i}{d_i}} = \left( { - 80} \right) + \left( { - 240} \right) + 0 + 320 + 400 = 400\]
\[\sum {{f_i} = 4 + 24 + 20 + 32 + 20 = 100} \]
Now putting the values in formula we get,
\[
mean = a + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \\
\Rightarrow 135 + \dfrac{{400}}{{100}} \\
\Rightarrow 135 + 4 \\
\Rightarrow 139 \\
\]
Thus the mean height of students is \[ \Rightarrow 139cm\].
Note:
We can go with our traditional formula of finding the mean but that will make our work too tedious so used this assumed mean method. Also note that when you find the term \[{f_i}{d_i}\] we should be aware of negative signs also. Because that will affect when we find the summation.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

