
Apply step deviation method to find arithmetic mean (rounded off to nearest integer) of the following frequency distribution.
Variate 5 10 15 20 25 30 frequency 20 43 75 67 72 45
Variate | 5 | 10 | 15 | 20 | 25 | 30 |
frequency | 20 | 43 | 75 | 67 | 72 | 45 |
Answer
503.1k+ views
Hint:
Here we have to find the mean by step deviation method. For that, we will first assume the mean from the given and then we will find the value of difference of consecutive variates. Then we will subtract assumed mean from each variate and divide it by the difference of variates obtained, then we will put all the values obtained in the formula to get the required mean by step deviation method.
Complete step by step solution:
We know the formula of mean by step deviation method.
$\overline{x}=A+h\left[ \dfrac{1}{N}\sum{{{f}_{i}}{{u}_{i}}} \right]$
A= assumed mean
h = difference between the classes
f i= frequency
${{u}_{i}}=\dfrac{{{x}_{i}}-A}{h}$
$N=\sum{{{f}_{i}}}$
Let’s assume mean to be 20. i.e. $A=20$
Here, $h=10-5=5$
Let’s draw the table and find the values one by one.
Now, we will find the sum of ${{f}_{i}}{{u}_{i}}$ and sum of${{f}_{i}}$ .
Sum of ${{f}_{i}}=20+43+75+67+72+45$
After addition, we get
Sum of ${{f}_{i}}=20+43+75+67+72+45$
Sum of ${{f}_{i}}=322$
Now, we will find the sum of ${{f}_{i}}{{u}_{i}}$
Sum of ${{f}_{i}}{{u}_{i}}=-60-86-75+0+72+90$
On addition, we get
Sum of ${{f}_{i}}{{u}_{i}}=-59$
Now, we will put the values in the formula of mean.
$\overline{x}=20+5\left[ \dfrac{1}{322}\times -59 \right]$
On further simplification, we get
$ \overline {x}=20-0.91 \\
\Rightarrow \overline {x}=19.09 $
Thus, the required mean is 19.09.
Note:
We need to know the meaning of following terms
1) A frequency distribution is defined as a tabular representation of data that represents the number of observations within a given class interval. The class intervals should always be mutually exclusive. Frequency distribution is used for summarizing and for getting overview of categorical data.
2) Mean is also known as average and it is defined as the ratio of sum of all numbers to the total number of numbers.
Here we have to find the mean by step deviation method. For that, we will first assume the mean from the given and then we will find the value of difference of consecutive variates. Then we will subtract assumed mean from each variate and divide it by the difference of variates obtained, then we will put all the values obtained in the formula to get the required mean by step deviation method.
Complete step by step solution:
We know the formula of mean by step deviation method.
$\overline{x}=A+h\left[ \dfrac{1}{N}\sum{{{f}_{i}}{{u}_{i}}} \right]$
A= assumed mean
h = difference between the classes
f i= frequency
${{u}_{i}}=\dfrac{{{x}_{i}}-A}{h}$
$N=\sum{{{f}_{i}}}$
Let’s assume mean to be 20. i.e. $A=20$
Here, $h=10-5=5$
Let’s draw the table and find the values one by one.
VARIATE(xi) | FREQUENCY(fi) | Deviation$ & ={{x}_{i}}-A \\ & ={{x}_{i}}-20 \\ $ | ${{u}_{i}}=\dfrac{{{x}_{i}}-20}{5}$ | ${{f}_{i}}{{u}_{i}}$ |
5 | 20 | -15 | $\dfrac{-15}{5}=-3$ | $20\times -3=-60$ |
10 | 43 | -10 | $\dfrac{-10}{5}=-2$ | $43\times -2=-86$ |
15 | 75 | -5 | $\dfrac{-5}{5}=-1$ | $75\times -1=-75$ |
20 | 67 | 0 | $\dfrac{0}{5}=0$ | $67\times 0=0$ |
25 | 72 | 5 | $\dfrac{5}{5}=1$ | $72\times 1=72$ |
30 | 45 | 10 | $\dfrac{10}{5}=2$ | $45\times 2=90$ |
Now, we will find the sum of ${{f}_{i}}{{u}_{i}}$ and sum of${{f}_{i}}$ .
Sum of ${{f}_{i}}=20+43+75+67+72+45$
After addition, we get
Sum of ${{f}_{i}}=20+43+75+67+72+45$
Sum of ${{f}_{i}}=322$
Now, we will find the sum of ${{f}_{i}}{{u}_{i}}$
Sum of ${{f}_{i}}{{u}_{i}}=-60-86-75+0+72+90$
On addition, we get
Sum of ${{f}_{i}}{{u}_{i}}=-59$
Now, we will put the values in the formula of mean.
$\overline{x}=20+5\left[ \dfrac{1}{322}\times -59 \right]$
On further simplification, we get
$ \overline {x}=20-0.91 \\
\Rightarrow \overline {x}=19.09 $
Thus, the required mean is 19.09.
Note:
We need to know the meaning of following terms
1) A frequency distribution is defined as a tabular representation of data that represents the number of observations within a given class interval. The class intervals should always be mutually exclusive. Frequency distribution is used for summarizing and for getting overview of categorical data.
2) Mean is also known as average and it is defined as the ratio of sum of all numbers to the total number of numbers.
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