NCERT Solution Class 10th Maths Chapter – 13 Statistics Examples

Example 1. The marks obtained by 30 students of Class 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, with the corresponding frequency f. So, let us put them in a column as shown in Table 13.1.

Now, x = Σf_ix_i/Σf_i = 1779/30 = 59.3

Therefore, the mean marks obtained is 59.3.

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.

Solution: Let us find the class marks, x, 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 = x – 50/10

We now find d_i and u_i and put them in Table 13.7.

Solution: Here, the class size varies, and the x’s are large. Let us still apply the step-
deviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8.

So,u= -160\45 Therefore, x= 200+20 {-160\450} =200- 47.11= 152.89

This tells us that, on an average, the number of wickets taken by these 45 bowlers
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 10 cricket matches are as follows:
Find the mode of the data.

2     6     4    5    0     2    1     3     2     3 
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:

mode = ∫ + [ƒ_i -ƒ₀\2ƒ_i -ƒ_{0 – ^ƒ2\-]× h}

where ∫ = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
ƒ_i = frequency of the modal class,
ƒ₀ = frequency of the class preceding the modal class,
_^ƒ2 = frequency of the class succeeding the modal class.
Let us consider the following examples to illustrate the use of this formula.

Find the mode of this data.

Solution: Here the maximum class frequency is 8, and the class corresponding frequency is 3-5. So, the modal class is 3-5.

Now

modal class = 3-5, lower limit (∫) of modal class = 3, class size (h) = 2
frequency (ƒ_i) of the modal class = 8,
frequency (ƒ₀) of class preceding the modal class = 7,
frequency (_^ƒ2) of class succeeding the modal class = 2.
Now, let us substitute these values in the formula:

mode = ∫ + [ƒ_i -ƒ₀\2ƒ_i -ƒ_{0 – ƒ2\-]× h}

= 3+ (8-7\2×8-7-2) ×2=3+2\7 =3.286

Therefore, the mode of the data above is 3.286.

Example 6. The marks distribution of 30 students in a mathematics examination given in Table 13.3 of Example 1. Find the mode of this data. Also compar interpret the mode and the mean.

Solution: Refer to Table 13.3 of Example 1. Since the maximum number of stu
(i.e., 7) have got marks in the interval 40 – 55, the modal class is 40 – 55. There

the lower limit (∫) of the modal class = 40,
the class size (h) = 15,
the frequency (ƒ_i) of modal class = 7,
the frequency (ƒ₀) of the class preceding the modal class = 3,
the frequency (ƒ₂) of the class succeeding the modal class= 6.
Now, using the formula:

Mode  = ∫+ [ƒ_i -ƒ₀\2ƒ_i -ƒ_{0 -ƒ 2\-]× h}

Mode  = 40 + (7-3\14 – 6 – 3) ×15 = 52
we get
tc.
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 ave
student obtained 62 marks

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,…, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150,…, 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, n/2 = 51/2 = 25.5 This observation lies in the class 145 – 150. Then,
∫ (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.

Using the formula, Median = ∫ + [(n/2 – cf)/f] x h, we have
Median 145 + [25.5 – 11/18] x 5
= 145 + 72.5/18 = 149.03
So, the median height of the girls is 149.03 cm.
This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height.

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, ∫ = 500, f = 20, cf = 36 + x, h = 100
Using the formula: Median = ∫ + [(n/2 – cf)/f] h, we get
525 = 500 + [50 – 36 – x/20] x 100
i.e., 525 – 500 + (14 -x) x 5
i.e., 25 = 70 – 5x
i.e., 5x = 70 – 25 = 45
so, x = 9
Therefore, from (1), we get 9+ y = 24
y = 15