# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5 | Set 2

### Question 11. Find the mean, median, and mode of the following data:

Classes |
0-50 |
50-100 |
100-150 |
150-200 |
200-250 |
250-300 |
300-350 |

Frequency |
2 |
3 |
5 |
6 |
5 |
3 |
1 |

**Solution:**

Let mean (A) = 175

Classes

Class Marks

(x)

Frequency

(f)c.f.

d_{i}= x – A

A = 175f_{i}* d_{i}0-50 25 2 2 -150 -300 50-100 75 3 5 -100 -300 100-150 125 5 10 -50 -250 150-200 175-A 6 16 0 0 200-250 225 5 21 50 250 250-300 275 3 24 100 300 300-350 325 1 25 150 150 Total25-150

Find Median:Here N = 25, 5/3 = 25/2 = 12.5 or 13, it lies in the class interval = 50-200.

l = 150, F = 10, f = 6, h = 50

Using median formula, we get

= 150 + 20.83

= 170.83

Find Mean:Using mean formula we get

= 175 – 6

= 169

Find Mode:Using mode formula we get

= 150 + 25

= 175

### Question 12. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.

Number of cars |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |

Frequency |
7 |
14 |
13 |
12 |
20 |
11 |
15 |
8 |

**Solution:**

From the given table we conclude that

Modal class = 40-50 (it has maximum frequency)

Also,

l = 40, f = 20, f

_{1}= 12, f_{2}= 11 and h = 10By using mode formula, we get

= 40 + 4.70

= 44.7

### Question 13. 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:

Monthly consumption(in units) |
65-85 |
85-105 |
105-125 |
125-145 |
145-165 |
165-185 |
185-205 |

No. of consumers |
4 |
5 |
13 |
20 |
14 |
8 |
4 |

**Solution:**

Let mean (A) = 135

Monthly consumptionClass Marks (x)No. of consumers (f)c.f.d = x – Af.d65-85 75 4 4 -60 -240 85-105 95 5 9 -40 -200 105-125 115 13 22 -20 -260 125-145 135 20 42 0 0 145-165 155 14 56 20 280 165-185 175 8 64 40 320 185-205 195 4 68 60 240 Total68140

Find Median:Here, N = 34

N/2 = 34,

Class interval = 25-145

Also,

l = 125, F = 22, f = 20 and h = 20

By using the median formula, we get

= 125 + 12

= 137 units

Find Mean:By using the mean formula, we get

Mean =

= 135 + 2.05

= 137.05 units

Find Mode:By using the mode formula, we get

= 125 + 10.76

= 135.76 units

### Question 14. 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:

Number of letters |
1-4 |
4-7 |
7-10 |
10-13 |
13-16 |
16-19 |

Number of surnames |
6 |
30 |
40 |
16 |
4 |
4 |

### Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.

**Solution:**

Let mean (A) = 8.5

Number of lettersClass Marks(x)No. of surnames (f)c.f.d = x- Af.d.1-4 2.5 6 6 -6 -36 4-7 5.5 30 36 -3 -90 7-10 8.5-A 40 76 0 0 10-13 11.5 16 92 3 48 13-16 14.5 4 96 6 24 16-19 17.5 4 100 9 36 Total100-18

Find Median:Here, N = 100

So, N/2 = 50

Class interval = 7-10

l = 7, F = 36, f = 40 and h =3

By using the median formula, we get

= 7 + 1.05

= 8.05

Find Mean:By using the mean formula, we get

Mean =

= 8.5 + 0.18

= 8.32

Find Mode:We have,

N = 100

N/2 = 100/2 = 50

Here, the cumulative frequency is just greater than N/2 = 76,

Hence, the median class = 7 – 10

l = 7, h = 10 – 7 = 3, f = 40, F = 36

By using the mode formula, we get

Mode = l +

= 7 +

= 7 + 30/34

= 7 + 0.88

= 7.88

### Question 15. Find the mean, median, and mode of the following data:

Class |
0 – 20 |
20 – 40 |
40 – 60 |
60 – 80 |
80 – 100 |
100 – 120 |
120 – 140 |

Frequency |
6 |
8 |
10 |
12 |
6 |
5 |
3 |

**Solution:**

Class intervalMid valueFrequency (f)fxCumulative frequency0 – 20 10 6 60 6 20 – 40 30 8 240 17 40 – 60 50 10 500 24 60 – 80 70 12 840 36 80 – 100 90 6 540 42 100 – 120 110 5 550 47 120 – 140 130 3 390 50 N = 50âˆ‘fx = 3120

Find Mean:By using the mean formula, we get

Mean =

Find Median:We have,

N = 50

Then, N/2 = 50/2 = 25

Here, the cumulative frequency just greater than N/2 = 36

Hence, the median class = 60 – 80

l = 60, h = 80 – 60 = 20, f = 12, F = 24

By using the median formula, we get

Median = l +

= 60 +

= 60 + 20/12

= 60 + 1.67

= 61.67

Find Mode:We have,

The maximum frequency = 12

Model class = 60 – 80

l = 60, h = 80 – 60 = 20, f = 12, f

_{1}= 10, f_{2}= 6By using the mode formula, we get

Mode = l +

= 60 +

= 60 + 40/8

= 65

### Question 16. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure |
Frequency |
Expenditure |
Frequency |

1000 – 1500 |
24 |
3000 – 3500 |
30 |

1500 – 2000 |
40 |
3500 – 4000 |
22 |

2000 – 2500 |
33 |
4000 – 4500 |
16 |

2500 – 3000 |
28 |
4500 – 5000 |
7 |

**Solution:**

From the given table we conclude that

The maximum class frequency = 40

So, modal class = 1500 – 2000

l = 1500, f = 40, h = 500, f

_{1 }= 24, f_{2 }= 33By using the mode formula, we get

Mode = l +

= 1500 +

= 1500 +

= 1500 + 347.826

= 1847.826 â‰ˆ 1847.83

Hence, the modal monthly expenditure = Rs. 1847.83

Now we will find class marks as

Class mark =

Class size (h) of given data = 500

Let mean(a) = 2750, now we are going to calculate d

_{i}u_{i}as follows:

Expenditure (In Rs)Number of families f_{i}X_{i}d_{i}= x_{i}– 2750U_{i}f_{i}u_{i}1000 – 1500 24 1250 -1500 -3 -72 1500 – 2000 40 1750 -1000 -2 -80 2000 – 2500 33 2250 -500 -1 -33 2500 – 3000 28 2750 0 0 0 3000 – 3500 30 3250 500 1 30 3500 – 4000 22 3750 1000 2 44 4000 – 4500 16 4250 1500 3 48 4500 – 5000 7 4750 2000 4 28 Total200-35From the table we conclude that

âˆ‘f

_{i }= 200âˆ‘f

_{i}d_{i }= -35Mean = a +

= 2750 +

= 2750 – 87.5

= 2662.5

Hence, the mean monthly expenditure = Rs. 2662.5

### Question 17. The given distribution shows the number of runs scored by some top batsmen of the world in one day international cricket matches.

Runs scored |
No. of batsmen |
Runs scored |
No. of batsmen |

3000 – 4000 |
4 |
7000 – 8000 |
6 |

4000 – 5000 |
18 |
8000 – 9000 |
3 |

5000 – 6000 |
9 |
9000 – 10000 |
1 |

6000 – 7000 |
7 |
10000 – 11000 |
1 |

### Find the mode of the data

**Solution:**

From the given table we conclude that

The maximum class frequency = 18

So, modal class = 4000 – 5000

and

l = 4000, f = 18, h = 1000, f

_{1 }= 4, f_{2 }= 9By using the mode formula, we get

Mode = l +

= 4000 +

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

Hence, the mode of given data = 4608.7 runs.

### Question 18. The frequency distribution table of agriculture holdings in a village is given below:

Area of land (in hectares): |
1 – 3 |
3 – 5 |
5 – 7 |
7 – 9 |
9 – 11 |
11 – 13 |

Number of families |
20 |
45 |
80 |
55 |
40 |
12 |

### Find the modal agriculture holdings of the village.

**Solution:**

From the given table we conclude that

The maximum class frequency = 80,

So, the modal class = 5-7

and

l = 5, f

_{0}= 45, h = 2, f_{1 }= 80, f_{2 }= 55By using the mode formula, we get

Mode = l +

= 5 +

= 5 +

= 5 +

= 5 + 1.2

= 6.2

So, the modal agricultural holdings of the village = 6.2 hectares.

### Question 19. The monthly income of 100 families are given as below:

Income in (in Rs) |
Number of families |

0 – 5000 |
8 |

5000 – 10000 |
26 |

10000 – 15000 |
41 |

15000 – 20000 |
16 |

20000 – 25000 |
3 |

25000 – 30000 |
3 |

30000 – 35000 |
2 |

35000 – 40000 |
1 |

### Calculate the modal income.

**Solution:**

From the given table we conclude that

The maximum class frequency = 41,

So, modal class = 10000-15000.

Here, l = 10000, f

_{1}= 41, f_{0}= 26, f_{2}= 16 and h = 5000Therefore, by using the mode formula, we get

Mode = l +

= 10000 +

= 10000 +

= 10000 +

= 10000 + 15 Ã— 125

= 10000 + 1875

= 11875

So, the modal income = Rs. 11875.

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