Examples - Chapter 13 Statistics class 10 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 13 - Statistics - Ncert Solutions class 10 - Maths

Example 1 :

The marks obtained by 30 students of Class $\mathrm{X}$ of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students.

Solution:

Recall that to find the mean marks, we require the product of each $x_i$ with the corresponding frequency $f_f$. So, let us put them in a column as shown in Table 13.1.

Now,
$
\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1779}{30}=59.3
$

Therefore, the mean marks obtained is 59.3 .
In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean.

Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say 15 . Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table (see Table 13.2).

Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is,
$
\text { Class mark }=\frac{\text { Upper class limit }+ \text { Lower class limit }}{2}
$

With reference to Table 13.2 , for the class $10-25$, the class mark is $\frac{10+25}{2}$, i.e., 17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table 13.3. These class marks serve as our $x_i$ 's. Now, in general, for the $i$ th class interval, we have the frequency $f_i$ corresponding to the class mark $x_i$. We can now proceed to compute the mean in the same manner as in Example 1.

The sum of the values in the last column gives us $\Sigma f_i x_i$. So, the mean $\bar{x}$ of the given data is given by
$
\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1860.0}{30}=62
$

This new method of finding the mean is known as the Direct Method.
We observe that Tables 13.1 and 13.3 are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact mean, while 62 an approximate mean.

Sometimes when the numerical values of $x_i$ and $f_i$ are large, finding the product of $x_i$ and $f_i$ becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations.

We can do nothing with the $f_i$ 's, but we can change each $x_i$ to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these $x_i$ 's? Let us try this method.

The first step is to choose one among the $x_i^{\prime}$ s as the assumed mean, and denote it by ' $a$ '. Also, to further reduce our calculation work, we may take ' $a$ ' to be that $x_i$ which lies in the centre of $x_1, x_2, \ldots, x_n$. So, we can choose $a=47.5$ or $a=62.5$. Let us choose $a=47.5$.

The next step is to find the difference $d_i$ between $a$ and each of the $x_i$ 's, that is, the deviation of ' $a$ ' from each of the $x_i$ 's.
i.e.,
$
d_i=x_i-a=x_i-47.5
$

The third step is to find the product of $d_i$ with the corresponding $f_i$, and take the sum of all the $f_i d_i$ 's. The calculations are shown in Table 13.4.

So, from Table 13.4, the mean of the deviations, $\bar{d}=\frac{\Sigma f_i d_i}{\Sigma f_i}$.
Now, let us find the relation between $\bar{d}$ and $\bar{x}$.
Since in obtaining $d_i$, we subtracted ' $a$ ' from each $x_i$, so, in order to get the mean $\bar{x}$, we need to add ' $a$ ' to $\bar{d}$. This can be explained mathematically as:
Mean of deviations,
So,
$
\begin{aligned}
\bar{d} & =\frac{\Sigma f_i d_i}{\Sigma f_i} \\
\bar{d} & =\frac{\Sigma f_i\left(x_i-a\right)}{\Sigma f_i} \\
& =\frac{\Sigma f_i x_i}{\Sigma f_i}-\frac{\Sigma f_i a}{\Sigma f_i} \\
& =\bar{x}-a \frac{\Sigma f_i}{\Sigma f_i} \\
& =\bar{x}-a \\
\bar{x} & =a+\bar{d} \\
\bar{x} & =a+\frac{\Sigma f_i d_i}{\Sigma f_i}
\end{aligned}
$

Substituting the values of $a, \Sigma f_i d_i$ and $\Sigma f_i$ from Table 13.4, we get
$
\bar{x}=47.5+\frac{435}{30}=47.5+14.5=62 .
$

Therefore, the mean of the marks obtained by the students is 62 .
The method discussed above is called the Assumed Mean Method.

Example 2 :

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Source : Seventh All India School Education Survey conducted by NCERT
Solution :

Let us find the class marks, $x_i$, of each class, and put them in a column (see Table 13.6):

Here we take $a=50, h=10$, then $d_i=x_i-50$ and $u_i=\frac{x_i-50}{10}$.
We now find $d_i$ and $u_i$ and put them in Table 13.7.

From the table above, we obtain $\Sigma f_i=35, \Sigma f_i x_i=1390$,
$
\Sigma f_i d_i=-360, \quad \Sigma f_i \mu_i=-36 .
$

Using the direct method, $\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1390}{35}=39.71$
Using the assumed mean method,
$
\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=50+\frac{(-360)}{35}=39.71
$

Using the step-deviation method,
$
\bar{x}=a+\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h=50+\left(\frac{-36}{35}\right) \times 10=39.71
$

Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71 .
Remark : The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of $x_i$ and $f_i$. If $x_i$ and $f_i$ are sufficiently small, then the direct method is an appropriate choice. If $x_i$ and $f_i$ are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and $x_i$ are large numerically, we can still apply the step-deviation method by taking $h$ to be a suitable divisor of all the $d_i$ 's.

Example 3 :

The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Solution :

Here, the class size varies, and the $x_i \mathrm{~s}$ are large. Let us still apply the stepdeviation method with $a=200$ and $h=20$. Then, we obtain the data as in Table 13.8.

So, $\bar{u}=\frac{-106}{45}$. Therefore, $\bar{x}=200+20\left(\frac{-106}{45}\right)=200-47.11=152.89$.
This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89 .
Now, let us see how well you can apply the concepts discussed in this section!

Example 4:

The wickets taken by a bowler in cricket matches are as follows: 

Find the mode of the data.
Solution :

Let us form the frequency distribution table of the given data as follows:

Clearly, 2 is the number of wickets taken by the bowler in the maximum number (i.e., 3) of matches. So, the mode of this data is 2.

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:
$
\text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h
$
where $l=$ lower limit of the modal class,
$h=$ size of the class interval (assuming all class sizes to be equal),
$f_1=$ 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.

Let us consider the following examples to illustrate the use of this formula.
Example 5 :

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Find the mode of this data.
Solution :

Here the maximum class frequency is 8 , and the class corresponding to this frequency is $3-5$. So, the modal class is $3-5$.
Now
modal class $=3-5$, lower limit $(l)$ of modal class $=3$, class size $(h)=2$
frequency $\left(f_1\right)$ of the modal class $=8$,
frequency $\left(f_0\right)$ of class preceding the modal class $=7$,
frequency $\left(f_2\right)$ of class succeeding the modal class $=2$.

Now, let us substitute these values in the formula :

$
\begin{aligned}
\text { Mode } & =l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =3+\left(\frac{8-7}{2 \times 8-7-2}\right) \times 2=3+\frac{2}{7}=3.286
\end{aligned}
$

Therefore, the mode of the data above is 3.286 .

Example 6 :

The marks distribution of 30 students in a mathematics examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean.
Solution :

Refer to Table 13.3 of Example 1. Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55 , the modal class is $40-55$. Therefore, the lower limit $(l)$ of the modal class $=40$, the class size $(h)=15$, the frequency $\left(f_1\right)$ of modal class $=7$, the frequency $\left(f_0\right)$ of the class preceding the modal class $=3$, the frequency $\left(f_2\right)$ of the class succeeding the modal class $=6$.

Now, using the formula:
$
\begin{aligned}
& \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h, \\
& \text { Mode }=40+\left(\frac{7-3}{14-6-3}\right) \times 15=52
\end{aligned}
$
we get
So, the mode marks is 52 .
Now, from Example 1, you know that the mean marks is 62 .
So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks.
Remarks :
1. In Example 6, the mode is less than the mean. But for some other problems it may be equal or more than the mean also.
2. It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most of the students. In the first situation, the mean is required and in the second situation, the mode is required.

Example 7:

A survey regarding the heights (in $\mathrm{cm}$ ) of 51 girls of Class $\mathrm{X}$ of a school was conducted and the following data was obtained:

Find the median height.
Solution :

To calculate the median height, we need to find the class intervals and their corresponding frequencies.

The given distribution being of the less than type, $140,145,150, \ldots, 165$ give the upper limits of the corresponding class intervals. So, the classes should be below 140, $140-145,145-150, \ldots, 160-165$. Observe that from the given distribution, we find that there are 4 girls with height less than 140 , i.e., the frequency of class interval below 140 is 4 . Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval $140-145$ is $11-4=7$. Similarly, the frequency of $145-150$ is $29-11=18$, for $150-155$, it is $40-29=11$, and so on. So, our frequency distribution table with the given cumulative frequencies becomes:

Now $n=51$. So, $\frac{n}{2}=\frac{51}{2}=25.5$. This observation lies in the class $145-150$. Then,
$l$ (the lower limit) $=145$,
cf (the cumulative frequency of the class preceding $145-150)=11$,
$f$ (the frequency of the median class $145-150$ ) $=18$,
$h$ (the class size) $=5$.
$
\begin{aligned}
& \text { Using the formula, Median }=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) \times h \text {, we have } \\
& \text { Median }=145+\left(\frac{25.5-11}{18}\right) \times 5 \\
& =145+\frac{72.5}{18}=149.03 \text {. } \\
&
\end{aligned}
$

So, the median height of the girls is $149.03 \mathrm{~cm}$.
This means that the height of about $50 \%$ of the girls is less than this height, and $50 \%$ are taller than this height.
Example 8:

The median of the following data is 525. Find the values of $x$ and $y$, if the total frequency is 100.

Solution : 

It is given that $n=100$
So, $76+x+y=100$, i.e., $x+y=24$
The median is 525 , which lies in the class $500-600$
So, $l=500, f=20, \quad \mathrm{cf}=36+x, \quad h=100$

Using the formula :
$
\begin{aligned}
\text { Median } & =l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) h, \text { we get } \\
525 & =500+\left(\frac{50-36-x}{20}\right) \times 100
\end{aligned}
$
i.e.,
$
\begin{aligned}
525-500 & =(14-x) \times 5 \\
25 & =70-5 x \\
5 x & =70-25=45 \\
x & =9
\end{aligned}
$

Therefore, from (1), we get $9+y=24$
i.e.,
$
y=15
$

Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement.

The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance.

However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2 , and the five others have frequency $20,25,20$, 21,18 , then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data.

In problems where individual observations are not important, and we wish to find out a 'typical' observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency.

In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc.

Remarks :
1. There is a empirical relationship between the three measures of central tendency:
$
3 \text { Median }=\text { Mode }+2 \text { Mean }
$
2. The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here.