# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1

**Question 1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.**

Expenditure (in rupees) (x) | Frequency (f_{i}) | Expenditure (in rupees) (x_{i}) | Frequency (f_{i}) |

100 – 150 | 24 | 300 – 350 | 30 |

150 – 200 | 40 | 350 – 400 | 22 |

200 – 250 | 33 | 400 – 450 | 16 |

250 – 300 | 28 | 450 – 500 | 7 |

**Find the average expenditure (in rupees) per household.**

**Solution:**

Let the assumed mean (A) = 275

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Class intervalMid value (x_{i})d_{i }= x_{i}– 275u_{i}= (x_{i}– 275)/50Frequency f_{i}f_{i}u_{i}100 – 150 125 -150 -3 24 -72 150 – 200 175 -100 -2 40 -80 200 – 250 225 -50 -1 33 -33 250 – 300 275 0 0 28 0 300 – 350 325 50 1 30 30 350 – 400 375 100 2 22 44 400 – 450 425 150 3 16 48 450 – 500 475 200 4 7 28 N = 200Σ f_{i}u_{i }= -35It’s seen that A = 275 and h = 50

So,

Mean = A + h x (Σf

_{i }u_{i}/N)= 275 + 50 (-35/200)

= 275 – 8.75

= 266.25

**Question 2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.**

Number of plants: | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 | 12 – 14 |

Number of house: | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

**Which method did you use for finding the mean, and why?**

**Solution:**

From the given data,

To find the class interval we know that,

Class marks (x

_{i}) = (upper class limit + lower class limit)/2Now, let’s compute x

_{i}and f_{i}x_{i}by the following

Number of plantsNumber of house (f_{i})x_{i}f_{i}x_{i}0 – 2 1 1 1 2 – 4 2 3 6 4 – 6 1 5 5 6 – 8 5 7 35 8 – 10 6 9 54 10 – 12 2 11 22 12 – 14 3 13 39 TotalN = 20Σ f_{i}u_{i }= 162Here,

Mean = Σ f

_{i}u_{i}/N= 162/ 20

= 8.1

Thus, the mean number of plants in a house is 8.1

We have used the direct method as the values of class mark x

_{i}and f_{i}is very small.

**Question 3. Consider the following distribution of daily wages of workers of a factory**

Daily wages (in ₹) | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |

Number of workers: | 12 | 14 | 8 | 6 | 10 |

**Find the mean daily wages of the workers of the factory by using an appropriate method.**

**Solution:**

Let us assume mean (A) = 150

Class intervalMid value x_{i}d_{i }= x_{i}– 150u_{i}= (x_{i }– 150)/20Frequency f_{i}f_{i}u_{i}100 – 120 110 -40 -2 12 -24 120 – 140 130 -20 -1 14 -14 140 – 160 150 0 0 8 0 160 – 180 170 20 1 6 6 180 – 200 190 40 2 10 20 N= 50Σ f_{i}u_{i }= -12It’s seen that,

A = 150 and h = 20

So,

Mean = A + h x (Σf

_{i }u_{i}/N)= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

**Question 4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute 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: | 65 – 68 | 68 – 71 | 71 – 74 | 74 – 77 | 77 – 80 | 80 – 83 | 83 – 86 |

Number of women: | 2 | 4 | 3 | 8 | 7 | 4 | 2 |

**Solution:**

Using the relation (x

_{i}) = (upper class limit + lower class limit)/ 2And, class size of this data = 3

Let the assumed mean (A) = 75.5

So, let’s calculate d

_{i}, u_{i}, f_{i}u_{i}as following:

Number of heart beats per minuteNumber of women (f_{i})x_{i}d_{i}= x_{i}– 75.5u_{i}= (x_{i }– 755)/hf_{i}u_{i}65 – 68 2 66.5 -9 -3 -6 68 – 71 4 69.5 -6 -2 -8 71 – 74 3 72.5 -3 -1 -3 74 – 77 8 75.5 0 0 0 77 – 80 7 78.5 3 1 7 80 – 83 4 81.5 6 2 8 83 – 86 2 84.5 9 3 6 N = 30Σ f_{i}u_{i }= 4From table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σf

_{i }u_{i}/N)= 75.5 + 3 x (4/30

= 75.5 + 2/5

= 75.9

Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.

**Question 5. Find the mean of each of the following frequency distributions:**

Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |

Frequency: | 6 | 8 | 10 | 9 | 7 |

**Solution:**

Let’s consider the assumed mean (A) = 15

Class intervalMid – value x_{i}d_{i }= x_{i }– 15u_{i }= (x_{i }– 15)/6f_{i}f_{i}u_{i}0 – 6 3 -12 -2 6 -12 6 – 12 9 -6 -1 8 -8 12 – 18 15 0 0 10 0 18 – 24 21 6 1 9 9 24 – 30 27 12 2 7 14 N = 40Σ f_{i}u_{i }= 3From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σf

_{i }u_{i}/N)= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

**Question 6. Find the mean of the following frequency distribution:**

Class interval: | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 | 130 – 150 | 150 – 170 |

Frequency: | 18 | 12 | 13 | 27 | 8 | 22 |

**Solution:**

Let’s consider the assumed mean (A) = 100

Class intervalMid – value x_{i}d_{i }= x_{i }– 100u_{i }= (x_{i }– 100)/20f_{i}f_{i}u_{i}50 – 70 60 -40 -2 18 -36 70 – 90 80 -20 -1 12 -12 90 – 110 100 0 0 13 0 110 – 130 120 20 1 27 27 130 – 150 140 40 2 8 16 150 – 170 160 60 3 22 66 N= 100Σ f_{i}u_{i }= 61From the table it’s seen that,

A = 100 and h = 20

Mean = A + h x (Σf

_{i }u_{i}/N)= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

**Question 7. Find the mean of the following frequency distribution:**

Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 6 | 7 | 10 | 8 | 9 |

**Solution:**

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf

_{i }u_{i}/N)= 20 + 8 x (7/40)

= 20 + 1.4

= 21.4

**Question 8. Find the mean of the following frequency distribution:**

Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |

Frequency: | 7 | 5 | 10 | 12 | 6 |

**Solution:**

Let’s consider the assumed mean (A) = 15

Class intervalMid – value x_{i}d_{i }= x_{i }– 15u_{i }= (x_{i }– 15)/6f_{i}f_{i}u_{i}0 – 6 3 -12 -2 7 -14 6 – 12 9 -6 -1 5 -5 12 – 18 15 0 0 10 0 18 – 24 21 6 1 12 12 24 – 30 27 12 2 6 12 N = 40Σ f_{i}u_{i }= 5From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σf

_{i }u_{i}/N)= 15 + 6 x (5/40)

= 15 + 0.75

= 15.75

**Question 9. Find the mean of the following frequency distribution:**

Class interval: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Frequency: | 9 | 12 | 15 | 10 | 14 |

**Solution:**

Let’s consider the assumed mean (A) = 25

Class intervalMid – value x_{i}d_{i }= x_{i }– 25u_{i }= (x_{i }– 25)/10f_{i}f_{i}u_{i}0 – 10 5 -20 -2 9 -18 10 – 20 15 -10 -1 12 -12 20 – 30 25 0 0 15 0 30 – 40 35 10 1 10 10 40 – 50 45 20 2 14 28 N = 60Σ f_{i}u_{i }= 8From the table it’s seen that,

A = 25 and h = 10

Mean = A + h x (Σf

_{i }u_{i}/N)= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

**Question 10. Find the mean of the following frequency distribution:**

Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 5 | 9 | 10 | 8 | 8 |

**Solution:**

Let’s consider the assumed mean (A) = 20

Class intervalMid – value x_{i}d_{i }= x_{i }– 20u_{i }= (x_{i }– 20)/8f_{i}f_{i}u_{i}0 – 8 4 -16 -2 5 -10 8 – 16 12 -4 -1 9 -9 16 – 24 20 0 0 10 0 24 – 32 28 4 1 8 8 32 – 40 36 16 2 8 16 N = 40Σ f_{i}u_{i }= 5From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf

_{i }u_{i}/N)= 20 + 8 x (5/40)

= 20 + 1

= 21

**Question 11. Find the mean of the following frequency distribution:**

Class interval: | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 5 | 6 | 4 | 3 | 2 |

**Solution:**

Let’s consider the assumed mean (A) = 20

Class intervalMid – value x_{i}d_{i }= x_{i }– 20u_{i }= (x_{i }– 20)/8f_{i}f_{i}u_{i}0 – 8 4 -16 -2 5 -12 8 – 16 12 -8 -1 6 -8 16 – 24 20 0 0 4 0 24 – 32 28 8 1 3 9 32 – 40 36 16 2 2 14 N = 20Σ f_{i}u_{i }= -9From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf

_{i }u_{i}/N)= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

**Question 12. Find the mean of the following frequency distribution:**

Class interval: | 10 – 30 | 30 – 50 | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 |

Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |

**Solution:**

Let’s consider the assumed mean (A) = 60

Class intervalMid – value x_{i}d_{i }= x_{i }–60u_{i }= (x_{i }– 60)/20f_{i}f_{i}u_{i}10 – 30 20 -40 -2 5 -10 30 – 50 40 -20 -1 8 -8 50 – 70 60 0 0 12 0 70 – 90 80 20 1 20 20 90 – 110 100 40 2 3 6 110 – 130 120 60 3 2 6 N = 50Σ f_{i}u_{i }= 14From the table it’s seen that,

A = 60 and h = 20

Mean = A + h x (Σf

_{i }u_{i}/N)= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

**Question 13. Find the mean of the following frequency distribution:**

Class interval: | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 |

Frequency: | 6 | 10 | 8 | 12 | 4 |

**Solution:**

Let’s consider the assumed mean (A) = 50

Class intervalMid – value x_{i}d_{i }= x_{i }– 50u_{i }= (x_{i }– 50)/10f_{i}f_{i}u_{i}25 – 35 30 -20 -2 6 -12 35 – 45 40 -10 -1 10 -10 45 – 55 50 0 0 8 0 55 – 65 60 10 1 12 12 65 – 75 70 20 2 4 8 N = 40Σ f_{i}u_{i }= -2From the table it’s seen that,

A = 50 and h = 10

Mean = A + h x (Σf

_{i }u_{i}/N)= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5