Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.1

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.

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

Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

Solution:

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

!! Since f_{i} and x_{i} are of small values we use direct method.

Question 2.

Consider the following distribution of daily wages of 50 workers of a factory. (AS_{4})

Daily wages in Rupees | 200 – 250 | 250 – 300 | 300 – 350 | 350 – 400 | 400- 450 |

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:

Here, the x_{i} are of large numerical values in 250 – 300. So, a = 275.

So we use Assumed Mean method then,

\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

Here, the Assumed mean is taken as 275.

∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = 275 + \(\frac{1900}{50}\) = 275 + 38 = 313

Question 3.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. (AS_{4})

Daily pocket allowance(in Rupees) | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |

Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |

Solution:

\(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

\(\overline{\mathrm{x}}\) = 18 (given)

⇒ 18 = \(\frac{752+20 f}{(44+f)}\)

⇒ 18(44 + f) = 752 + 20f ⇒ 20f – 18f = 792 – 752

⇒ 2f = 40

∴ f = \(\frac{40}{2}\) = 20

Question 4.

Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method. (AS_{4})

Number of heart beats/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:

\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

= 75.5 + \(\frac{12}{30}\) = 75.5 + 0.4 = 75.9

Question 5.

In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges. (AS_{4})

Number of oranges | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 |

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

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose ?

Solution:

Here, we use step division method where a = 22, h = 5

\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h

⇒ 22 + \(\frac{25}{400}\) × 5

⇒ 22 + 0.31 = 22.31

Question 6.

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

Daily expenditure (in Rupees) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |

Number of house holds | 4 | 5 | 12 | 2 | 2 |

Find the mean daily expenditure on food by a suitable method. (AS_{4})

Solution:

Here, a = 225, h = 50

\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h

= 225 + \(\frac{(-7)}{25}\) × 50 = 225 – 14 = 211

The average daily expenditure on food = ₹ 211

Question 7.

To find out the concentration of SO_{2} in the air (in parts per million, i.e., ppm), the data was collect for 30 localities in a certain city and is presented below.

Concentration of SO_{2} in ppm |
0.00-0.04 | 0.04-0.08 | 0.08-0.12 | 0.12-0.16 | 0.16-0.20 | 0.20-0.24 |

Frequency | 4 | 9 | 9 | 2 | 4 | 2 |

Find the mean concentration of SO_{2} in the air. (AS_{4})

Solution:

∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

= \(\frac{2.96}{30}\) = 0.0986666 …….

≅ 0.099 ppm

Question 8.

A class teacher has the following attendence record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term. (AS_{4})

Number of days | 35-38 | 38-41 | 41-44 | 44-47 | 47-50 | 50-53 | 53-56 |

Number of students | 1 | 3 | 4 | 4 | 7 | 10 | 11 |

Solution:

Here, a = 51.5

∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

= 51.5 – \(\frac{99}{40}\)

⇒ 51.5 – 2.475 = 49.025 ≅ 49 days

Question 9.

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (AS_{4})

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

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

Solution:

Here a = 70, h = 10

∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h

⇒ \(\overline{\mathrm{x}}\) = 70 – \(\frac{2}{35}\) × 10

= 70 – \(\frac{20}{35}\) = 70 – 0.5714

= 69.4285 ≅ 69.43%