Practice the AP 10th Class Maths Bits with Answers Chapter 14 Statistics on a regular basis so that you can attempt exams with utmost confidence.

## AP State Syllabus 10th Class Maths Bits 14th Lesson Statistics with Answers

Question 1.

If mean of 8, 6, 4, x, 3, 6 and ‘0’ is ‘4’, then find the value of x.

Answer:

1

Explanation:

\(\frac{6+8+4+x+3+6+0}{7}\) = 4

⇒ 27 + x = 28 ⇒ x = 1

Question 2.

Where the extreme values of some data influences high?

Answer:

In AM

Question 3.

In a data ‘n’ scores are given and if ‘n’ is odd, then find median.

Answer:

(\(\frac{n+1}{2}\))^{th} event

Question 4.

Find the class mark of 10 – 25.

Answer:

17.5

Question 5.

Mode of the data 5, 3, 4, – 2, 3, 2, 2, 1, p is 3, then the value of ‘p’.

Answer:

3

Question 6.

Find the class interval of the class 11 – 20.

Answer:

10

Question 7.

Mean of 1,2, x, 3 is ‘0’, then the value of ‘x’.

Answer:

-6

Question 8.

If the sum of 15 observations is 420, then find their mean.

Answer:

28

Explanation:

Mean = \(\frac{420}{15}\) = 28

Question 9.

Find the mid value of the class 10 – 19.

Answer:

14.5

Question 10.

Find the median of 2, 3, 4, 5, 6, 7.

Answer:

4.5

Explanation:

Median = \(\frac{4+5}{2}\) = 4.5

Question 11.

To elect the leader of your class from 3 contestants, which central measures is to be considered ?

Answer:

Mode

Question 12.

From the given graph of ogives, find median.

Answer:

40

Question 13.

Find median of the scores

1, 3, 5, 7, 9, ……………. 99.

Answer:

50

Question 14.

Find the mode of the values sin 0°, cos 0°, sin 90° and tan 45°.

Answer:

1

Explanation:

Mode in the values at 0, 1, 1, 1 is 1.

Question 15.

Find the mean of first four odd prime number.

Answer:

6.5

Question 16.

Find A.M. of x – 5, x, x + 5.

Answer:

x

Question 17.

Mode of 3, 4, 5 and x is 5, then find x.

Answer:

5

Explanation:

x = 5

Question 18.

Find the mode of the data

5, 6, 9, 10, 6, 11, 4, 6, 10, 4.

Answer:

6

Question 19.

The letter that represents \(\frac{\mathbf{x}_{\mathbf{i}}-\mathbf{a}}{\mathbf{h}}\)

which is used in measuring mean is …………….

Answer:

u_{i}

Question 20.

In “more than ogive curve” we consider in drawing …………………..

Answer:

More than cumulative frequency, lower limits.

Question 21.

Observe the following tables.

For finding Arithmetic Mean by Direct method, the suggested frequency distribution table is ……………….

Answer:

only (1) is true.

Question 22.

If \(\overline{\mathbf{x}}\), is the mean of x_{1} x_{2}, x_{3}, ……………… x_{n}

(n times), then find \(\sum_{i=1}^{n}\)(x_{1} – \(\overline{\mathbf{x}}\))

Answer:

0

Question 23.

Mode can be calculated by = l + (\(\frac{\mathbf{f}_{1}-\mathbf{f}_{0}}{2 \mathbf{f}_{1}-\mathbf{f}_{0}-\mathbf{f}_{2}}\)) × h

here f_{1} represents ……………..

Answer:

Frequency of the modal class.

Question 24.

Find the x – coordinate of the point of intersection of the two ogives of grouped data.

Answer:

Median of the data.

Question 25.

3, 2, 4, 3, 5, 2, x, 6. If the mode of this data is 3, then find ‘x’.

Answer:

3

Question 26.

For the terms, x + 1, x + 2, x – 1, x + 3 and x – 2 (x G N), if the median of the data is 12, then find x.

Answer:

11

Question 27.

Which one of the following is NOT a measure of central tendency?

Mean, Median, Mode, Range

Answer:

Range

Question 28.

Find the most stable measure of central tendency.

Answer:

Mean

Question 29.

Find mode of 2004, 2005, 2006, ……………

2019.

Answer:

No mode

Question 30.

Mode = 24.5, Mean = 29.75, then find median.

Answer:

28

Question 31.

Unimodal data may have ……………… modes.

Answer:

1

Question 32.

Who is known as father of statistics ?

Answer:

Fisher

Question 33.

Find AM of \(\frac{1}{3}\),\(\frac{7}{12}\),\(\frac{3}{4}\),\(\frac{1}{2}\),\(\frac{5}{6}\)

Answer:

\(\frac{3}{5}\)

Question 34.

Find the mean of first 5 odd multiples of 5.

Answer:

25

Explanation:

\(\overline{\mathbf{x}}\) = \(\frac{5+15+25+35+45}{5}\) = \(\frac{125}{5}\) = 25

Question 35.

Find the mean of 6,-4,\(\frac{2}{3}\),\(\frac{5}{4}\),\(\frac{7}{6}\)

Answer:

\(\frac{11}{20}\)

Question 36.

In the figure, find the value of median of the data using the graph of less than ogive and more than ogive.

Answer:

20

Question 37.

Find mode of the following distribution.

Answer:

52

Question 38.

Find the measure of central tendency which take into account all data terms.

Answer:

Mean

Question 39.

6, 3, 5, 6, 7, 5, 8, 7, 6, 2k + 1, 9, 7, 13. If the mode of this data is 7, then find ‘k’

Answer:

3

Explanation:

Mode = 7, so 2k + 1 = 7 ⇒ k = 3

Question 40.

For a given data with 50 observations ‘the less than ogive’ and the more than ogive intersect at (15.5, 20). Find the median of the data.

Answer:

15.5

Question 41.

Find median of first 8 prime numbers.

Answer:

9

Question 42.

In a data mean = 72.5 and median = 73.9, then find mode.

Answer:

76.7

Explanation:

Mode = 3 Median – 2 Mean

= 3 × 73.9 – 2 × 72.5

= 221.7 – 145 = 76.7

Question 43.

How much the sum of all deviations taken from AM?

Answer:

0

Question 44.

Find the modal class in the following frequency distribution.

Class | Frequency |

0 – 10
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 |
3
9 15 30 18 5 |

Answer:

30 – 40

Question 45.

Mean – mode is equal to

Answer:

3 (Mean – Median)

Question 46.

In an arranged series of an even num¬ber 2n terms write the median.

Answer:

\(\frac{1}{2}\)(n^{th} and (n + 1)^{th} term)

Question 47.

In a data maximum value = x, minimum value = y, then find range.

Answer:

x – y

Consider the following frequency distribution.

Monthly income | Number of families |

More than or equal to 10000 More than or equal to 13000 More than or equal to 16000 More than or equal to 19000 More than or equal to 22000 |
100
85 69 50 33 |

Question 48.

Find the number of families having income range from ₹ 16000 to ₹ 19000.

Answer:

19

Question 49.

C.I of 1 – 10 is …………….

Answer:

10

Question 50.

In calculating mode Δ_{1} is equal to

Answer:

f – f_{1}

Question 51.

|| represents ………………

Answer:

7

Question 52.

Find mode of any 3 consecutive numbers.

Answer:

No mode.

Question 53.

Find the mean of the following data.

Answer:

8.1

Question 54.

Find AM of first n odd numbers.

Answer:

n

Question 55.

The class mark of 10 – 25 is ……………..

Answer:

17.5

Question 56.

Mean of 5,7,9,x is 9, then find ‘x’.

Answer:

15

Question 57.

Find AM of a – 2, a, a + 2.

Answer:

a

Question 58.

Cumulative frequency curves are called as ………………. curves.

Answer:

Ogive

Question 59.

Find the class marks of a class interval.

Answer:

\(\frac{\text { Upper boundary + lower boundary }}{2}\)

Question 60.

Write the abscissa of the point of intersection of the ‘less than’ type and ‘more than type’ cumulative frequency curves of a grouped data.

Answer:

Mode

Question 61.

Find AM of 1^{2}, 2^{2}, 3^{3}, 4^{2}, …………….. , 20^{2}.

Answer:

Question 62.

Write mode of first ‘n’ natural numbers.

Answer:

No mode.

Question 63.

Find mid value of the class 10 – 20.

Answer:

15

Question 64.

Mid values are used to calculate ……………

Answer:

Mean

Question 65.

Find mode of 1, 2, 3, ……………. 10, 10.

Answer:

10

Question 66.

Find the mean of the following data.

Answer:

15.1

Question 67.

Find the median of the data 5, 3, 10, 7, 2, 9, 11,2, 6.

Answer:

6

Question 68.

Histograiri’consists of ………………..

Answer:

Rectangles

Question 69.

Write empirical relation among mean, median and mode.

Answer:

Mode = 3 median – 2 mean

Question 70.

Write the class marks of a class x – y.

Answer:

\(\frac{x+y}{2}\)

Question 71.

Find mean of 1, 2, 3, …………… n.

Answer:

\(\frac{n+1}{2}\)

Question 72.

Find AM of 23, 24, 24, 22, 10.

Answer:

22.6

Question 73.

Find the modal class for the following distribution.

Answer:

30 – 40

Explanation:

Maximum frequency is 57,

i.e., 30 – 40 class.

Question 74.

For a given data with 60 observations, ‘the less than’ ogive and ‘the more than ogive’ intersect at (66.5, 30). Find the median of the data.

Answer:

66.5

Question 75.

Data having two modes is called ……………….

Answer:

Bimodal

Question 76.

pie diagram consists of …………….

Answer:

Sectors

Question 77.

Write the Range of first 5 natural num-bers.

Answer:

4

Question 78.

If the mean of 10, 12, 18, 13, P and 17 is 15, then find ‘P’.

Answer:

20

Explanation:

\(\frac{10+12+18+13+\mathrm{P}+17}{6}\) = 15

⇒ 70 + P = 90 ⇒ P = 20

Question 79.

Find range of 1,2,3, …………….. 10.

Answer:

9

Question 80.

For a distribution with odd numbers (n) of observations, find the median is ………………. observation.

Answer:

\(\frac{n+1}{2}\)^{th}

Question 81.

If each observation of a data is increased by ‘a’, then mean is increases by ……………….

Answer:

a

Question 82.

Find mode of 5, 6, 9,10,6,12,3,6,11, 10, 4, 6, 7.

Answer:

6

Question 83.

……………….. is effected by extreme values.

Answer:

Mean

Question 84.

1-8, 9-16, 17-24, ………………… then find C.I.

Answer:

8

Explanation:

9 – 1 = 8

Question 85.

Mode is the value of variate which occurs …………….. number of times.

Answer:

Maximum

Question 86.

Find the mean of first five prime numbers.

Answer:

5.6

Question 87.

Find mean of -8, -4 and 4, 8.

Answer:

0

Question 88.

Mean of n-observations x_{1}, x_{2}, …………….. x_{n} repeated , f_{1},f_{2}, f_{3}, ……………. , f_{n} times is

Answer:

Question 89.

Find mean of 7, 6, 5, 0, 7, 8, 9.

Answer:

6

Question 90.

Representing the data with the help of pictures is called

Answer:

pictograph

Question 91.

If the mean of the data 2, a + 1, a, a – 2 is 4, then find a.

Answer:

5

Explanation:

\(\overline{\mathbf{x}}\) = \(\frac{2+a+1+a+a-2}{4}\) = 4

⇒ 3a + 1 = 16 ⇒ a = 5

Question 92.

£f(x) = 200, n=20, then find \(\overline{\mathbf{x}}\).

Answer:

10

\(\overline{\mathbf{x}}\) = \(\frac{\Sigma \mathrm{fx}}{\mathrm{n}}=\frac{200}{20}\) = 10

Question 93.

Median = 52.5, Mean = 54, then find Mode.

Answer:

49.5

Question 94.

In the formula of mode in the grouped ……………… data l represents

Answer:

Lower limit of the class with highest frequency.

Question 95.

Cumulative frequency is used to calculate in ……………….

Answer:

Median

Question 96.

Find the lower limit of the modal class of the following data.

Answer:

10

Question 97.

Construction of cumulative frequency table is useful in determining the ………………

Answer:

Median

Question 98.

\(\frac{\text { Sum of observations }}{\text { Number of observations }}\) is equal to

Answer:

Mean

Question 99.

If mode of a distribution is 8 and its mean is 8, then find median.

Answer:

8

Question 100.

Find mode of 0, 1, 2, 3, 3, 3, 7.

Answer:

3

Question 101.

For a distribution with even number (n) of observation, the median is ……………….

Answer:

Question 102.

If the mean of x_{1} x_{2}, …………….. x_{n} is \(\overline{\mathbf{x}}\), then find the mean of \frac{\mathbf{x}_{1}}{\mathbf{a}}, \(\frac{\mathbf{x}_{2}}{\mathbf{a}}\) ……………….. \(\frac{\mathbf{x}_{n}}{\mathbf{a}}\)

Answer:

\(\frac{\overline{\mathrm{x}}}{\mathrm{a}}\)

Question 103.

Find AM of 3 and 4.

Answer:

3.5

Question 104.

For a given data with 120 observations, the “less than ogive” and the “more than ogive” intersect at (42.5,60). Find the median of the data.

Answer:

42.5

Question 105.

Find the AM of 10 consecutive num¬bers starting with n + 1.

Answer:

x + 5.5

Question 106.

………………. is based on all observations.

Answer:

Mean

Question 107.

Mode of a continuous grouped distribution is ………………….

Answer:

l + \(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\) × h

Question 108.

Find mode of 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3.

Answer:

3,7

Question 109.

Find the sum of lower limit of median class and upper limit of modal class is ………………

Answer:

90

Question 110.

The information collected is called ……………..

Answer:

data

Question 111.

Find AM of 1, 2, x, 3 is 0, then find ‘x’.

Answer:

-6

Question 112.

For a symmetrical distribution, which is correct?

A) Mean < Mode < Median B) Mean > Mode > Median

C) Mode = \(\frac{\text { Mean }+\text { Median }}{2}\)

D) Mean = Median = Mode

Answer:

D) Mean = Median = Mode

Question 113.

\(\overline{\mathbf{x}}\) = 2p + q, M = p + 2q, then find Z.

Answer:

4q – p

Question 114.

Find the median class of the following distribution.

Answer:

30 – 40

Question 115.

A data has 13 observations arranged in descending order which observation represents the median of data?

Answer:

7^{th}

Question 116.

For a continuous grouped frequency distribution, the median is given by

Answer:

l + (\(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\)) × h

Question 117.

A data arrange in descending order has 25 observations. Which observation represents the median?

Answer:

13^{th}

Question 118

……………… of all bars is same in bar graph.

Answer:

Width

Question 119.

Find mean of a + 1, a + 3, a + 4 and a + 8.

Answer:

a + 4

Explanation:

Mean = \(\frac{a+1+a+3+a+4+a+8}{4}\)

= \(\frac{4 a+16}{4}\) = a + 4

Question 120.

If assumed mean is ‘a’, then write the mean.

Answer:

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

Question 121.

Find the upper limit of median class of the following distribution.

Answer:

18

❖ Choose the correct answer satisfying the following statements.

Question 122.

Statement (A) : The arithmetic mean of the following given frequency distribution table is 13.81.

Statement (B) : \(\overline{\mathbf{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i) Both A and B are true.

Explanation:

Both A and B are true, B is the correct explanation of the A.

Hence, (i) is the correct option.

Question 123.

Statement (A) : If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is 30.

Statement (B) : Median = \(\frac{(n+1)^{t h}}{2}\)

value, if n is odd.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

iii) A is false, B is true.

Explanation:

Arranging the terms in ascending order, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52

Median value = (\(\frac{11+1}{2}\))^{th} = 6^{th} value = 27

∴ Option (iii) is true.

Question 124.

Statement (A) : If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.

Statement (B) :

Median = \(\frac{1}{2}\)(mode + 2 mean)

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

ii) A is true, B is false.

Explanation:

Median = \(\frac { 1 }{ 3 }\) (Mode + 2 Mean)

= \(\frac { 1 }{ 3 }\) (60 + 2 × 66) = 64

∴ Option (ii) is correct.

Read the below passages and answer to the following questions.

The following table gives the weekly wages of workers in a factory.

Question 125.

Find the mean.

Answer:

69

Explanation:

Mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{5520}{80}\) = 69

Question 126.

Find the modal class.

Answer:

55 – 60

Explanation:

Modal Class : We know that class of maximum frequency is called the modal class, i.e., 55 – 60 is the modal class.

Question 127.

Find the number of workers getting weekly wages, below ₹ 80.

Answer:

60

The marks of 20 students in a test were as follows : 5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20.

Explanation:

Number of workers getting weekly wages below ₹ 80 according to table = 60 workers.

Question 128.

Calculate the mean.

Answer:

13

Explanation:

Arrange in ascending order

Question 129.

Calculate the median.

Answer:

13.5

Explanation:

Here, n = 20 which is an even number

= 13.5

Question 130.

Calculate the mode.

Answer:

15

A professor keeps data on students tabulated by performance and sex of the student. The data is kept on a computer disk.

Explanation:

In the data, 15 occurs the maximum times i.e., 3 times.

∴ Mode = 15

Question 131.

How many students are both female and excellent?

Answer:

0

Explanation:

There is no female excellent student in the class.

Question 132.

What proportion of good students are male?

Answer:

0.73

Explanation:

Proportion of good male students

= \(\frac{22}{30}\) = 0.73

Question 133.

What proportion of female students are good?

Answer:

0.26

Explanation:

Proportion of good female students

\(\frac{8}{30}\) = 0.26

Question 134.

Write the short form of the expansion in symbolic form.

1 + 2 + 3 + 4 + ……………………. + n

Answer:

Σn