## CBSE Class 10 Maths Chapter-14 Statistics – Free PDF Download

Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 14 – Statistics prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.

CBSE Class 10 Maths Chapter-14 Statistics Important Questions

**CBSE Class 10 Maths Important Questions Chapter 14 – Statistics**

**3 Mark Questions**

**1. The following table shows the weekly wages drawn by number of workers in a factory, find the median of the following data.**

Weekly wages (in Rs.) | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |

No. of workers | 40 | 39 | 34 | 30 | 45 |

**Ans.** We have

Weekly wages (in Rs.) | No. of workers (f) | C.F |

0-100 | 49 | 40 |

100-200 | 39 | 79 |

200-300 | 34 | 113 |

300-400 | 30 | 143 |

400-500 | 45 | 188 |

Now and this is in 200-300 class.

Median class= 200-300

Here,

We know that

**2. Find the median of the following data:**

Marks | Frequency |

Less than 10 | 0 |

Less than 30 | 10 |

Less than 50 | 25 |

Less than 70 | 43 |

Less than 90 | 65 |

Less than 110 | 87 |

Less than 130 | 96 |

Less than 150 | 100 |

**Ans.** First of all we shall change cumulating series into simple series.

We have

X | F | C.F |

0-10 | 0 | 0 |

10-30 | 10 | 10 |

30-50 | 15 | 25 |

50-70 | 18 | 43 |

70-90 | 22 | 65 |

90-110 | 22 | 87 |

110-130 | 9 | 96 |

130-150 | 4 | 100 |

Now , which lies in 70-90 class

Median class = 70-90

Here,

We know that Median, Me =

**3. Find the median of the following data.**

Wages (in rupees) | No. of workers |

More than 150 | Nil |

More than 140 | 12 |

More than 130 | 27 |

More than 120 | 60 |

More than 110 | 105 |

More than 100 | 124 |

More than 90 | 141 |

More than 80 | 150 |

**Ans.** First of all we shall find simple frequencies.

Wages (in Rupees) (X) | No. of workers (F) | C.F |

80-90 | 9 | 9 |

90-100 | 17 | 26 |

100-110 | 19 | 45 |

110-120 | 45 | 90 |

120-130 | 33 | 123 |

130-140 | 15 | 138 |

140-150 | 2 | 150 |

Now which lies in 110-120 class

Median class = 110-120

Here,

We know that Me =

**4. Draw a less than Ogive for the following frequency distribution.**

Marks | No. of students |

0-4 | 4 |

4-8 | 6 |

8-12 | 10 |

12-16 | 8 |

16-20 | 4 |

**Ans.** We have

Marks | Frequency (F) | C.F |

0-4 | 4 | 4 |

4-8 | 6 | 10 |

8-12 | 10 | 20 |

12-16 | 8 | 28 |

16-20 | 4 | 32 |

Upper class limits | 4 | 8 | 12 | 16 | 20 |

Cumulative frequency | 4 | 10 | 20 | 28 | 32 |

Plot the points | (4,4) | (8,10) | (12,20) | (16,28) | (20,32) |

Join these points by a free hand curve. We get the required Ogivewhich is as follows:

**5. Find the mean age in years from the frequency distribution given below:**

Age (in yrs) | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | Total |

Frequency | 3 | 12 | 21 | 15 | 5 | 4 | 2 | 3 |

**Ans.** We have

Class-Interval | Mid value | |||

15-19 | 17 | 3 | -3 | -9 |

20-24 | 22 | 13 | -2 | -26 |

25-29 | 27 | 21 | -1 | -21 |

30-34 | 32 | 15 | 0 | 0 |

35-39 | 37 | 5 | 1 | 5 |

40-44 | 42 | 4 | 2 | 8 |

45-49 | 47 | 2 | 3 | 6 |

Total |

Let assumed mean ‘a’ = 32, Here h = 5

We know that Mean

**6. Find the median of the following frequency distribution:**

Wages (in Rs.) | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 |

No. of Laborers | 3 | 5 | 20 | 10 | 6 |

**Ans.** We have

Wages (in Rs.) | No. of laborers (f) | C.F |

200-300 | 3 | 3 |

300-400 | 5 | 8 |

400-500 | 20 | 28 |

500-600 | 10 | 38 |

600-700 | 6 | 44 |

Now and this lies in 400-500 class.

Median class = 400-500

Here,

We know that

**7. The following tables gives production yield per hectare of wheat of 100 farms of village:**

Production yield (in hr) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |

No. of farms | 2 | 8 | 12 | 24 | 38 | 16 |

**Change the distribution to a more than type distribution and draw its Ogive.**

**Ans.** More than type Ogive

Production yield (Kg/ha) | C.F |

More than or equal to 50 | 100 |

More than or equal to 55 | 98 |

More than or equal to 60 | 90 |

More than or equal to 65 | 78 |

More than or equal to 70 | 54 |

More than or equal to 75 | 16 |

Now, draw the Ogive by plotting the points (50,100), (55,98), (60,90), (65,78), (70,54), (75,16)

**8. The A.M of the following distribution is 47. Determine the value of P.**

Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |

Frequency | 8 | 15 | 20 | P | 5 |

**Ans.** We have

Class Interval | Mid-value | Frequency | |

0-20 | 10 | 8 | 80 |

20-40 | 30 | 15 | 450 |

40-60 | 50 | 20 | 1000 |

60-80 | 70 | P | 70P |

80-100 | 90 | 5 | 450 |

Since Mean,

Thus, P = 12

**9. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women choosing a suitable method.**

Number of heart beats per minute | No. of women |

65-68 | 2 |

68-71 | 4 |

71-74 | 3 |

74-77 | 8 |

77-80 | 7 |

80-83 | 4 |

83-86 | 2 |

**Ans.** Let assumed mean ‘a’ = 75.5. We have

No. of heart beats per minute | No. of women () | Class Mark i.e mid value | ||

65-68 | 2 | 66.5 | -3 | -6 |

68-71 | 4 | 69.5 | -2 | -8 |

71-74 | 3 | 72.5 | -1 | -3 |

74-77 | 8 | 75.5=a | 0 | 0 |

77-80 | 7 | 78.5 | 1 | 7 |

80-83 | 4 | 81.5 | 2 | 8 |

83-86 | 2 | 84.5 | 3 | 6 |

We know that

Mean [By step Deviation Method]

**10. Following distribution shows the marks obtained by a class of 100 students:**

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

Frequency | 10 | 15 | 30 | 32 | 8 | 5 |

**Change the distribution to less than type distribution and draw its Ogive.**

**Ans.** Less than type Ogive

Marks | Marks | Frequency | Cumulative Frequency |

10-20 | Less than 20 | 10 | 10 |

20-30 | Less than 30 | 15 | 25 |

30-40 | Less than 40 | 30 | 55 |

40-50 | Less than50 | 32 | 87 |

50-60 | Less than 60 | 8 | 95 |

60-70 | Less than 70 | 5 | 100 |

Now, draw the Ogiveby plotting (20,10), (30,25), (40,55), (50,87), (60,95), (70,100)

**11. Following table shows the daily pocket allowances given to the children of a multi-story building. The mean of the pocket allowances is Rs. 18. Find out the missing frequency.**

Class Interval | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |

Frequency | 3 | 6 | 9 | 13 | ? | 5 | 4 |

**Ans.** Let the missing frequency = f, we have

Class interval | Mid-value | f_{i}u_{i} | ||

11-13 | 3 | 12 | -3 | -9 |

13-15 | 6 | 14 | -2 | -12 |

15-17 | 9 | 16 | -1 | -9 |

17-19 | 13 | 18 | 0 | 0 |

19-21 | f | 20 | 1 | F |

21-32 | 5 | 22 | 2 | 10 |

23-25 | 4 | 24 | 3 | 12 |

Let assumed mean a = 18, Here h = 2

We know that mean

Hence, missing frequency = 8

**12. A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained. Find the median height.**

Height (in cm) | No. of girls |

Less than 140 | 4 |

Less than 145 | 11 |

Less than 150 | 29 |

Less than 155 | 40 |

Less than 160 | 46 |

Less than 165 | 51 |

**Ans.** We have,

Class Intervals | Frequency (f) | C.F |

Below 140 | 4 | 4 |

140-145 | 7 | 11 |

145-150 | 18 | 29 |

150-155 | 11 | 40 |

155-160 | 6 | 46 |

160-165 | 5 | 51 |

Here,which is in the class 145-150

Here,

Median

Median height of the girls = 149.03

**13. Calculate the mean for the following distribution:**

Class Interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 | 20-24 | 24-28 | 28-32 |

Frequency | 2 | 5 | 8 | 16 | 14 | 10 | 8 | 3 |

**Ans. **By stepdeviation Method

Let assumed mean a = 14

Class interval | Mid-value | Frequency | Deviation | Product |

0-4 | 2 | 2 | -12 | -24 |

4-8 | 6 | 5 | -8 | -40 |

8-12 | 10 | 8 | -4 | -32 |

12-16 | 14 | 16 | 0 | 0 |

16-20 | 18 | 14 | 4 | 56 |

20-24 | 22 | 10 | 8 | 80 |

24-28 | 26 | 8 | 12 | 96 |

28-32 | 30 | 3 | 16 | 48 |

Total |

We know that Mean

**14. The percentage of marks obtained by 100 students in an examination are given below:**

Marks | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 |

Frequency | 14 | 16 | 18 | 23 | 18 | 8 | 3 |

**Determine the median percentage of marks.**

**Ans.**

Class interval | Mid-value | Frequency | Deviation | Product |

0-4 | 2 | 2 | -12 | -24 |

4-8 | 6 | 5 | -8 | -40 |

8-12 | 10 | 8 | -4 | -32 |

12-16 | 14 | 16 | 0 | 0 |

16-20 | 18 | 14 | 4 | 56 |

20-24 | 22 | 10 | 8 | 80 |

24-28 | 26 | 8 | 12 | 96 |

28-32 | 30 | 3 | 16 | 48 |

Total |

We know that Mean

Therefore , which lies in the class 45-50

l_{1}(The lower limit of the median class) = 45

c(The cumulative frequency of the class preceding the median class) = 48

f(The frequency of the Median class)= 23

h(The class size) = 5

Median

So, the median percentage of marks is 45.4

**15. Draw a less than Ogive for the following frequency distribution:**

Marks | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |

No. of students | 4 | 6 | 10 | 8 | 4 |

**Ans. **We have

Marks | Frequency | C.F |

0-4 | 4 | 4 |

4-8 | 6 | 10 |

8-12 | 10 | 20 |

12-16 | 8 | 28 |

16-20 | 4 | 32 |

Upper class limits | 4 | 8 | 12 | 16 | 20 |

Cumulative Frequency | 4 | 10 | 20 | 28 | 32 |

Plot the points | (4, 4) | (8, 10) | (12, 20) | (16, 28) | (20, 32) |

Joint these points by a free hand curve; we get the required Ogive which is as follows:

**16. The A.M of the following frequency distribution is 53. Find the value of P.**

Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |

Frequency | 12 | 15 | 32 | P | 13 |

**Ans. **We have

Class Interval | Mid-value | Frequency | |

0-20 | 10 | 12 | 120 |

20-40 | 30 | 15 | 450 |

40-60 | 50 | 32 | 1600 |

60-80 | 70 | P | 70P |

80-100 | 90 | 13 | 1170 |

Since Mean

Thus, P = 28