
The number of students absent in a class was recorded for 120 days and the information is given in the following frequency table.
No. of students absent (x) 0 1 2 3 4 5 6 7 No. of days (f) 1 4 10 50 34 15 4 2
Find the mean number of students absent per day.
| No. of students absent (x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| No. of days (f) | 1 | 4 | 10 | 50 | 34 | 15 | 4 | 2 |
Answer
558.9k+ views
Hint: First, take each class as ${x_i}$ and frequency ${f_i}$. The mean value is equivalent to the fraction between the addition of a product of mid-value with frequency and the total frequency.
Complete step-by-step answer:
We are given the number of students absent.
Let us assume that ${f_i}$ represents the number of days and ${x_i}$ is the number of students absent.
Mean is the measure of the average of a set of values which can be calculated by dividing the sum of all the observations by the number of observations.
The frequency distribution table for the given data is as follows:
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{423}}{{120}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 3.525$
Hence the mean number of students absent per day is 3.525.
Note: In the mean formula, while computing $\sum {fx} $, don’t take the sum of $f$ and $x$ separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.
Complete step-by-step answer:
We are given the number of students absent.
Let us assume that ${f_i}$ represents the number of days and ${x_i}$ is the number of students absent.
Mean is the measure of the average of a set of values which can be calculated by dividing the sum of all the observations by the number of observations.
The frequency distribution table for the given data is as follows:
| ${x_i}$ | ${f_i}$ | ${f_i}{x_i}$ |
| 0 | 1 | 0 |
| 1 | 4 | 4 |
| 2 | 10 | 20 |
| 3 | 50 | 150 |
| 4 | 34 | 136 |
| 5 | 15 | 75 |
| 6 | 4 | 24 |
| 7 | 2 | 14 |
| $\sum {{f_i}} = 120$ | $\sum {{f_i}{x_i}} = 423$ |
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{423}}{{120}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 3.525$
Hence the mean number of students absent per day is 3.525.
Note: In the mean formula, while computing $\sum {fx} $, don’t take the sum of $f$ and $x$ separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

