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NCERT Solutions for Class 10 Maths Chapter 14- Statistics are curated by our team of subject experts at Toppr in a detailed manner. NCERT Solutions for statistics class 10 are made strictly in accordance with the CBSE Curriculum and the exam pattern. The NCERT textbook questions are answered in a way to provide you with a better understanding of the concepts in a systematic and step-by-step manner. It also contains the appropriate diagrams to explain the concepts to the students. As Class 10 exams are Board exams, these solutions will not only help the students in preparing for the board exams but also for the Olympiads. NCERT Solutions provided by Toppr are the best study material to excel in the exams. Also, the MCQs and long and short questions are all answered according to the weightage and the exam pattern. With the help of NCERT Statistics Class 10 Solutions you can also test your subject knowledge and analyze your shortcomings and work on them before the exams. These are the best resources designed after proper study and research and study to help the students in scoring good marks.

Table of Content

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.

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

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

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No. of plants | No. of houses $(f_{i})$ | Mid point $(X_{i})$ | $f_{i}x_{i}$ |

$0−2$ | $1$ | $1$ | $1$ |

$2−4$ | $2$ | $3$ | $6$ |

$4−6$ | $1$ | $5$ | $5$ |

$6−8$ | $5$ | $7$ | $35$ |

$8−10$ | $6$ | $9$ | $54$ |

$10−12$ | $2$ | $11$ | $22$ |

$12−14$ | $3$ | $13$ | $39$ |

Total | $20$ | $162$ |

Mean, $xˉ=n1 ∑f_{i}x_{i}$

Here, $n=20,∑f_{i}x_{i}=162$

Therefore, Mean, $xˉ=20162 =8.1$ plants

We have used direct method because numerical values of $f$ and $x$ are small.

Question 2

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

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

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

Daily wages(in rs) | Number of workers |

500-200 | 12 |

520-540 | 14 |

540-560 | 8 |

560-580 | 6 |

580-600 | 10 |

Solution

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Daily wages | $(fi)$ | $(xi)$ | $(fixi)$ |

$600−20$ | $12$ | $510$ | $6120$ |

$520−540$ | $14$ | $630$ | $7420$ |

$540−560$ | $8$ | $550$ | $4400$ |

$560−580$ | $6$ | $570$ | $3420$ |

$580−600$ | $10$ | $590$ | $$5900 |

$∑fi=50$ | $∑fixi⇒27260$ |

mean $∑fi∑fixi =5027260 =52726 =545.2$

$∴$ meam\n wage of $50$ employees is $545.2rs$

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 $f$.

Daily pocket allowance (in $Rs.$) | $11−13$ | $13−15$ | $15−17$ | $17−19$ | $19−21$ | $21−23$ | $23−25$ |

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

Solution

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Lets take assumed mean, $A$ as $16$

Class interval $=2$

$∴u_{i}=hx_{i}−A =2x_{i}−16 $

Now, we have $A=16,xˉ=18$ and $h=2$

We know that,

Mean, $xˉ=A+h(N1 ∑f_{i}u_{i})$

$⇒18=16+2(f+442f+24 )$

$⇒f+442f+24 =1$

$⇒f+44=2f+24$

$⇒f=20$

Hence, the value of missing frequency, $f=20$

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.

Number of heart beats per minute | $65−68$ | $68−71$ | $71−74$ | $74−77$ | $77−8$0 | $80−83$ | $83−86$ |

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

Solution

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Class interval | Mid value $(x_{i})$ | $(f_{i})$ | $u_{i}=3x_{i}−75.5 $ | $f_{i}u_{i}$ |

$65−68$ | $66.5$ | $2$ | $−3$ | $−6$ |

$68−71$ | $69.5$ | $4$ | $−2$ | $−8$ |

$71−74$ | $72.5$ | $3$ | $−1$ | $−3$ |

$74−77$ | $75.5$ | $8$ | $0$ | $0$ |

$77−80$ | $78.5$ | $7$ | $1$ | $7$ |

$80−83$ | $81.5$ | $4$ | $2$ | $8$ |

$83−86$ | $84.5$ | $2$ | $3$ | $6$ |

Total | $30$ | $4$ |

Lets take assumed mean, $A$ as $75.5$

Class interval $=3$

$∴u_{i}=hx_{i}−A =3x_{i}−75.5 $

$xˉ=A+h∑f_{i}∑f_{i}u_{i} =75.5+3×304 =75.5+0.4=75.9$

Hence, the mean heart beats per minute for these women is $75.9$

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.

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

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

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

Solution

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Number of mangoes | Number of boxes | $x_{i}$ | $f_{i}x_{i}$ |

50-52 | 15 | 51 | 765 |

53-55 | 110 | 54 | 5940 |

56-58 | 135 | 57 | 7695 |

59-61 | 115 | 60 | 6900 |

62-64 | 25 | 63 | 1575 |

$Σf_{i}=400$ | $Σf_{i}x_{i}=22875$ |

Hence, the mean number of mangoes kept in a packing box is $57.1875$.

Question 6

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

Find the mean daily expenses on food by a suitable method

Daily expenses (in Rs.) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |

No. of households | 4 | 5 | 12 | 2 | 2 |

Solution

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Daily Expense | Households (fi) | Mid Point (xi) | fi xi |

100 - 150 | 4 | 125 | 500 |

150 - 200 | 5 | 175 | 875 |

200 - 250 | 12 | 225 | 2700 |

250 - 300 | 2 | 275 | 550 |

300 - 350 | 2 | 325 | 650 |

25 | 5275 |

Question 7

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

Find the mean concentration of $SO_{2}$ in the air.

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

$0.00−0.04$ | $4$ |

$0.04−0.08$ | $9$ |

$0.08−0.12$ | $9$ |

$0.12−0.16$ | $2$ |

$0.16−0.20$ | $4$ |

$0.20−0.24$ | $2$ |

Solution

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Answer:-

$mean concentration ofSO_{2}in the air=Σf_{i}Σf_{i}x_{i} =302.96 =0.0986≈0.099$

Concentration of $SO_{2}$ (in ppm) | $x_{i}$ | Frequency $(f_{i})$ | $f_{i}x_{i}$ |

0.00-0.04 | 0.02 | 4 | 0.08 |

0.04-0.08 | 0.06 | 9 | 0.54 |

0.08-0.12 | 0.10 | 9 | 0.90 |

0.12-0.16 | 0.14 | 2 | 0.28 |

0.16-0.20 | 0.18 | 4 | 0.72 |

0.20-0.24 | 0.22 | 2 | 0.44 |

$Σf_{i}=30$ | $Σf_{i}x_{i}=2.96$ |

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.

Number of days: | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |

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

Solution

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Answer:-

The mean number of days a student was absent $=Σf_{i}Σf_{i}x_{i} =40499 =12.475$

No. of days | No. of student $f_{i}$ | $x_{i}$ | $f_{i}x_{i}$ |

0-6 | 11 | 3 | 33 |

6 - 10 | 10 | 8 | 80 |

10 - 14 | 7 | 12 | 84 |

14 - 20 | 4 | 17 | 68 |

20 - 28 | 4 | 24 | 96 |

28 - 38 | 3 | 33 | 99 |

38 - 40 | 1 | 39 | 39 |

$Σf_{i}=40$ | $Σf_{i}x_{i}=499$ |

Hence, the mean number of days a student was absent is $12.475.$

Question 9

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

Solution

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To find the class mark $(x_{i})$ for each interval, using the following relation:

$(x_{i})=2Upperclasslimit+Lowerclasslimit $

Class size $(h)$ for this data $=10$

Taking 70 as assumed mean $(a)$, calculating $d_{i},u_{i},f_{i}u_{i}$:

From the table:

$∑f_{i}=35$

$∑f_{i}u_{i}=−2$

Mean, $xˉ=a+(∑f_{i}∑f_{i}u_{i} )×h$

$=70+(35−2 )×10$

$=70−3520 $

$=70−74 $

$=70−0.57$

$=69.43$

Therefore, mean literacy rate is $69.43$%

$(x_{i})=2Upperclasslimit+Lowerclasslimit $

Class size $(h)$ for this data $=10$

Taking 70 as assumed mean $(a)$, calculating $d_{i},u_{i},f_{i}u_{i}$:

From the table:

$∑f_{i}=35$

$∑f_{i}u_{i}=−2$

Mean, $xˉ=a+(∑f_{i}∑f_{i}u_{i} )×h$

$=70+(35−2 )×10$

$=70−3520 $

$=70−74 $

$=70−0.57$

$=69.43$

Therefore, mean literacy rate is $69.43$%

Exercise 14.2

Question 1

The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Age (in years) | $5−15$ | $15−25$ | $25−35$ | $35−45$ | $45−55$ | $55−65$ |

Number of patients | $6$ | $11$ | $21$ | $23$ | $14$ | $5$ |

Solution

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In this case, class interval $35−45$ is the modal class.

Now,

Lower limit of modal class, $l=35,h=10,f_{1},=23,f_{0}=21,f_{2}=14$

We know that,

Mode $=l+(2f_{1}−f_{0}−f_{2}f_{1}−f_{0} )×h$

$=35+(2(23)−21−1423−21 )×10$

$=35+112 ×10$

$=35+1.818$

Mode $=36.8$

Lets take assumed mean, $A$ as $30$

Class interval $=10$

$∴u_{i}=hx_{i}−A =10x_{i}−30 $

$xˉ=A+h∑f_{i}∑f_{i}u_{i} =30+10×8043 $

$=30+5.375$$=35.375$

$≈35.37$

Hence, Mode $=36.8$ years, Mean $=35.37$ years.

Maximum number of patients admitted in the hospital are of age group $36.8$ years, while on an average the age of a patient admitted to the hospital is $35.37$ years.

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Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 15 : Probability

Question 1. State the empirical relationship between the Measures of Central Tendency.

Answer. The empirical relationship between the three measures of central tendency is as follows: 3 Median = Mode + 2 Mean

Question 2. What do you understand by Mean?

Answer. Mean is a measure of the central tendency of a distribution. It refers to the most common or the average value of the given data. There are two types of mean that can be calculated. One is the Arithmetic Mean and the other one is the Geometric Mean. The three methods to calculate the Arithmetic Mean are the direct method, the step-deviation method and the assumed mean method.

Question 3. What do you understand by Median?

Answer. Median refers to the middle number or value of the given set of data when the data is arranged in an ascending or descending order. When the data set contains odd numbers, the median will be the middle value or number that has half of the numbers above it and half of the number below it. However, in the case of even numbers, the median is calculated by adding the two middle numbers and dividing their sum by two. It is also a measure of central tendency.

Question 4. What do you understand by Mode?

Answer. Mode is also a measure of central tendency similar to the Mean and Median. Mode refers to the number or value that is most frequent in a given set of data. In other words, Mode is that value in a given data which occurs the most number of times. Thus, a distribution of data may have only one mode, more than one mode or no mode at all.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 15 : Probability