Well-designed AP 10th Class Maths Guide Chapter 13 Statistics Exercise 13.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Statistics Class 10 Exercise 13.1 Solutions – 10th Class Maths 13.1 Exercise Solutions

Question 1.

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

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

Solution:

We used direct method for finding the mean as the width of the class is very small and also the frequency for each class have very small values. So, it is easy to calculate.

∴ Mean=\(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{162}{20}\) = 8.1 plants.

Question 2.

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹) | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 |

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:

Note : Highest frequency class can be treated as the mean class.

\(\overline{\mathbf{x}}=\left(\frac{\sum f_i u_i}{\Sigma f_i}\right)\) × h

a = Assumed m£an class marks = 530

u_{i} = \(\frac{x_i-a}{h}\)

h = class height = 20

Σf_{i} = 50, Σf_{i}u_{i} = +48

Mean = 530 + \(\frac{48}{50}\) × 20 = 530 + 0.96 × 20 = 530 + 19.2

Therefore, mean = ₹ 549.2

Question 3.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Solution:

Note: Highest frequency class can be treated as the mean class.

\(\overline{\mathbf{x}}=\left(\frac{\sum f_i u_i}{\Sigma f_i}\right)\) × h

a = Assumed mean class marks = 18

u_{i} = \(\frac{x_i-a}{h}\)

h = class height = 2

x_{i} = mid-value

Σf_{i} = 44 + f, Σf_{i}u_{i} = -20 + f

Given Mean = 18

Mean \(\overline{\mathbf{x}}\) = 18 + \(\left(\frac{-20+\mathrm{f}}{44+\mathrm{f}}\right) \times 2\) = 18

\(\left(\frac{-20+\mathrm{f}}{44+\mathrm{f}}\right)\) = \(\frac{18-18}{2}\) = 0

So, -20 + f = 0

Therefore, f = 20

Question 4.

Thirty women were examined In a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

Mean = \(\overline{\mathbf{x}}\) = \(\frac{\sum f_i x_i}{\sum f_i}\)

x_{i} = class mid-value Σf_{i}x_{i} = 2277, Σf_{i} = Total frequency = 30

Mean \(\overline{\mathbf{x}}\) = \(\frac{2277}{30}\) = 75.9

Question 5.

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62 -64 |

Number of boxes | 15 | 110 | 135 | 115 | 25 |

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Note : Highest frequency class can be treated as the mean class.

Mean \(\overline{\mathbf{x}}\) = a + (\(\frac{\sum f_i u_i}{\sum f_i}\)) × h

a = Assumed mean class marks = 57

x_{i} = class mark

h = class height = 3

Σf_{i} = Total frequency = 400

Σf_{i}u_{i} = 25

Mean \(\overline{\mathbf{x}}\) = 57 + \(\frac{25}{400}\) × 3

= 57 + \(\frac{75}{400}\)

= 57 + 0.0625

Therefore, mean number of mangoes in each box is 57.0625

Question 6.

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in ₹) | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |

Number of households | 4 | 5 | 12 | 2 | 2 |

Find the mean daily expenditure on food by a suitable method.

Solution:

Note : Highest frequency class can be treated as the mean class.

Mean x = a + (\(\frac{\sum f_i u_i}{\sum f_i}\)) × h

a = class mark of the assumed mean class = 225

x_{i} = class mark

h = class height = 50

Σf_{i} = Total frequency = 25

Σf_{i}u_{i}= -7

Mean \(\overline{\mathbf{x}}\) = 225 + \(\frac{-7}{25}\) × 50

Mean \(\overline{\mathbf{x}}\) = 225- 14 = 211

Therefore, mean daily expenditure is ₹ 211

Question 7.

To find out the concentration of S02 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :

Concentration of SO_{2} (in ppm) |
Frequency |

0.00-0.04 | 4 |

0.04-0.08 | 9 |

0.08-0.12 | 9 |

0.12-0.16 | 2 |

0.16-0.20 | 4 |

0.20-0.24 | 2 |

find the mean concentration of SO_{2} in the air.

Solution:

Mean \(\overline{\mathbf{x}}\) = a + (\(\frac{\sum f_i u_i}{\sum f_i}\)) × h

a = class mark of the assumed mean class = 0.10

x_{i} = class mark,

h = height of the class = 0.04

Σf_{i} = total frequency = 30, Σf_{i}U_{i} = -1

Mean \(\overline{\mathbf{x}}\) = 0.10 + \(\frac{-1}{30}\) × 0.04

= 0.10 – \(\frac{0.04}{30}\) = 0.10 – 0.0013

Mean \(\overline{\mathbf{x}}\) = 0.0987 ppm

Therefore, concentration of S02 in air is 0.099 ppm. (Appro.)

Question 8.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

Mean \(\overline{\mathbf{x}}\) = \(\frac{\sum f_i x_i}{\sum f_i}\)

x_{i} = class mark

Σf_{i} = total frequency = 40

Σf_{i}x_{i} = 499

Mean \(\overline{\mathbf{x}}\) = \(\frac{499}{40}\) = 12.475 (or) 12.48 days

Therefore, mean number of days a student was absent = 12.48 days.

Question 9.

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |

Number of cities | 3 | 10 | 11 | 8 | 3 |

Solution:

Note : Highest frequency class can be treated as the mean class.

Mean \(\overline{\mathbf{x}}\) = a + (\(\frac{\sum f_i u_i}{\sum f_i}\)) × h

a = class mark of the assumed mean class = 70

xsub>i = class mark,

h = height of the class = 10

Σf_{i} = total frequency = 35, Σf_{i}u_{i} = -2

Mean \(\overline{\mathbf{x}}\) = 70 + (\(\frac{-2}{35}\)) × 10

= 70 – \(\frac{20}{35}\)

= 70-0.5714 = 69.428 (or) 69.43%.

Therefore, mean literacy rate = 69.43%