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## NCERT Class 10 Maths Textbook Chapter 14 With Answer Book PDF Free Download

### Chapter 14: Statistics

#### 14.1 Introduction

In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions.

You have also learned to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths), and frequency polygons.

In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median, and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median, and mode from ungrouped data to that of grouped data.

We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution, and how to draw cumulative frequency curves, called ogives.

#### 14.3 Mode of Grouped Data

Recall from Class IX, that a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.

Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data.

It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only.

#### 14.5 Graphical Representation of Cumulative Frequency Distribution

As we all know, pictures speak better than words. A graphical representation helps us in understanding given data at a glance.

In Class IX, we have represented the data through bar graphs, histograms, and frequency polygons. Let us now represent a cumulative frequency distribution graphically.

Author | NCERT |

Language | English |

No. of Pages | 35 |

PDF Size | 2.3 MB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### NCERT Solutions Class 10 Maths Chapter 14 Statistics

**1. A survey was conducted by a group of students as a part of their environment awareness program, 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 |

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

Solution:

In order to find the mean value, we will use the direct method because the numerical value of f_{i} and x_{i} are small.

Find the midpoint of the given interval using the formula.

Midpoint (x_{i}) = (upper limit + lower limit)/2

No. of plants(Class interval) | No. of house frequency (f_{i}) | Mid-point (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 |

Sum f_{i }= 20 | Sum f_{i}x_{i} = 162 |

The formula to find the mean is:

Mean = x̄ = ∑f_{i }x_{i} /∑f_{i }

= 162/20

= 8.1_{}

Therefore, the mean number of plants per house is 8.1

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

Daily wages (in Rs.) | 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:

Find the midpoint of the given interval using the formula.

Midpoint (x_{i}) = (upper limit + lower limit)/2

In this case, the value of the mid-point (x_{i}) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

So, u_{i }= (x_{i} – A)/h = u_{i }= (x_{i} – 150)/20

Substitute and find the values as follows:

Daily wages(Class interval) | Number of workersfrequency (f_{i}) | Mid-point (x_{i}) | u_{i }= (x_{i} – 150)/20 | f_{i}u_{i} |

100-120 | 12 | 110 | -2 | -24 |

120-140 | 14 | 130 | -1 | -14 |

140-160 | 8 | 150 | 0 | 0 |

160-180 | 6 | 170 | 1 | 6 |

180-200 | 10 | 190 | 2 | 20 |

Total | Sum f_{i }= 50 | Sum f_{i}u_{i} = -12 |

So, the formula to find out the mean is:

Mean = x̄ = A + h∑f_{i}u_{i} /∑f_{i }=150 + (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wage of the workers = Rs. 145.20

**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.**

Daily Pocket Allowance(in c) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-35 |

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

Solution:

To find out the missing frequency, use the mean formula.

Here, the value of mid-point (x_{i}) mean x̄ = 18

Class interval | Number of children (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i } |

11-13 | 7 | 12 | 84 |

13-15 | 6 | 14 | 84 |

15-17 | 9 | 16 | 144 |

17-19 | 13 | 18 = A | 234 |

19-21 | f | 20 | 20f |

21-23 | 5 | 22 | 110 |

23-25 | 4 | 24 | 96 |

Total | f_{i} = 44+f | Sum f_{i}x_{i} = 752+20f |

The mean formula is

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i }= (752+20f)/(44+f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752+20f)/(44+f)

⇒ 18(44+f) = (752+20f)

⇒ 792+18f = 752+20f

⇒ 792+18f = 752+20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

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

Number of heartbeats 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:

From the given data, let us assume the mean as A = 75.5

x_{i }= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the u_{i }and f_{i}u_{i} as follows:

Class Interval | Number of women (f_{i}) | Mid-point (x_{i}) | u_{i} = (x_{i} – 75.5)/h | f_{i}u_{i} |

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 | 0 | 0 |

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

80-83 | 4 | 81.5 | 3 | 8 |

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

Sum f_{i}= 30 | Sum f_{i}u_{i }= 4 |

Mean = x̄ = A + h∑f_{i}u_{i} /∑f_{i }

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

Therefore, the mean heartbeats per minute for these women is 75.9

**5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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:

Since the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between the two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval | Number of boxes (f_{i}) | Mid-point (x_{i}) | d_{i} = x_{i} – A | f_{i}d_{i} |

49.5-52.5 | 15 | 51 | -6 | 90 |

52.5-55.5 | 110 | 54 | -3 | -330 |

55.5-58.5 | 135 | 57 = A | 0 | 0 |

58.5-61.5 | 115 | 60 | 3 | 345 |

61.5-64.5 | 25 | 63 | 6 | 150 |

Sum f_{i} = 400 | Sum f_{i}d_{i} = 75 |

The formula to find out the Mean is:

Mean = x̄ = A +h ∑f_{i}d_{i} /∑f_{i }

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

NCERT Class 10 Maths Textbook Chapter 14 With Answer Book PDF Free Download