NCERT Solutions for Class 10 Maths Exercise 14.3 Chapter 14 Statistics- FREE PDF Download

NCERT Class 10 Maths Ch 14 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 14 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Statistics solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 14 – Statistics

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Ans. For Median:

Here, , then , which lies in interval 125 – 145.

Median class = 125 – 145

So, and

Now, Median =

= $125 + \frac{34 - 22}{20} \times 20$ = 125 + 12 = 137

For Mean:

From given data, Assume mean = 135, Width of the class = 20

Using formula, Mean = 135 + 20 (0.102)

= 135 + 2.04 = 137.04

For Mode:

In the given data, maximum frequency is 20 and it corresponds to the class interval 125 – 145.

Modal class = 125 – 145

And and

Mode =

= 125 + 10.76923

= 125 + 10.77

= 135.77

Hence, median, mean and mode of given data is 137 units, 137.04 units and 135.77 units.

2. If the median of the distribution given below is 28.5, then find the values of and

Ans.

Here, , then , also, median of the distribution is 28.5, which lies in interval 20 – 30.

Median class = 20 – 30

So, and

……….(i)

Now, Median =

Putting the value of in eq. (i), we get,

Hence the value of and are 8 and 7 respectively.

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.

Ans.

Here, , then , which lies in interval 35 – 40.

Median class = 35 – 40

So, and

Now, Median =

= 35 + 0.7575

= 35 + 0.76 (approx.)

= 35.76

Hence median age of given data is 35.76 years.

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.

Ans. Since the frequency distribution is not continuous, so firstly we shall make it continuous.

Here, , then , which lies in interval 144.5 – 153.5.

Median class = 144.5 – 153.5

So, and

Now, Median =

= 144.5 + 2.25

= 146.75

Hence median length of the leaves is 146.75 mm.

5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of the lamps.

(change the frequency of class interval 3000-3500 from 85 to 86)

Ans.

Here, , then , which lies in interval 3000 – 3500.

Median class = 3000 – 3500

So, and

Now, Median =

= 3000 + 406.9767441

= 3000 + 406.98 (approx.)

= 3406.98

Hence median life time of a lamp is 3406.98 hours.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: