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(i) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here the Number of observations are Even then Median = (28 + 29)/2 = 28.5

Median = 28.5 and n = 10

By using the formula to calculate Mean Deviation,

MD = 1/n ∑ⁿ_i=1|d_i|

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/10 × 47

= 4.7

∴ The Mean Deviation is 4.7.

(ii) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

34, 38, 43, 44, 47, 48, 53, 55, 63, 70

Here the Number of observations are Even then Median = (47 + 48)/2 = 47.5

Median = 47.5 and n = 10

By using the formula to calculate Mean Deviation,

MD = 1/n ∑ⁿ_i=1|d_i|

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

Correct option is: D) Midpoint

Mid -point is the representative of the whole class.

Correct option is: D) Midpoint

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation. 

Formula Used: Mean Deviation = \(\frac{\Sigma d_i}{n}\) 

Explanation: Here, Observations 22, 24, 30, 27, 29, 31, 25, 28, 41, 42are Given.

Since, Median is the middle number of all the observation, 

So, To Find the Median, Arrange the numbers in Ascending order, we get 

22, 24, 25, 27, 28, 29, 30, 31, 41, 42 

Here the Number of observations are Even then the middle terms are 28 and 29 

Therefore, The Median = \(\frac{28+29}{2}\) = 28.5

Deviation |d| = |x-Median| 

And, The number of observations is 10. 

Now, The Mean Deviation is

Hence, The Mean Deviation is 8.7

Mean Deviation = \(\frac{47}{10}\) = 4.7

Correct option is (C) 60

The lower limit of the class 60–69 is 60.

Correct option is  C) 60

Correct option is: A) 10

Correct option is: C) 15

Correct option is: B) median

Cumulative frequency are used in the calculations for median.

Correct option is: B) median

Correct option is: B) a

A.M (Algebraic men) = \(\frac {sum \, of \, observations}{Total \, number \, of\, observations}\)

= \(\frac {(a+2)+ a + (a-2)}{3} = \frac {3a}3 =a\)

Hence, A.M. of a+2, a, a-2 is a.

Correct option is: B) a

Correct option is (D) 14

The ascending order of observations is 6, 9, 13, 15, 19, 21.

Number of observations is n = 6.

\(\therefore\) Median \(=\frac{3^{rd}\text{ observation}+4^{th}\text{ observation}}2\)

\(=\frac{13+15}2\)

\(=\frac{28}2=14\)

Correct option is  D) 14

Correct option is: B) Range

Correct option is: D) 14.5

Mid value = \(\frac {Lower \,limit + upper \, limit}{2} \)

\(= \frac {10+19}2 = \frac {29}2 = 14.5\)

Correct option is: D) 14.5

Arrange the given data in ascending order

46, 48, 49, 64, 65, 66, 68, 72, 76, 79, 82, 82, 90, 91, 96, 100, 100

So from the data we know that

Minimum marks = 46 and Maximum marks = 100

We know that

Range of the given data = Maximum marks – Minimum marks = 100 – 46 = 54

Therefore, the range of the above data is 54.

No, infact the number of children who watched TV for 10 or more hour in a week is 4 + 2 i.e.,6.

Class of the class 90-120 = (Upper limit + Lower limit) / 2

So we get

Class of the class 90-120 = (90 + 120)/2 = 105

It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

Consider the lower limit of the class as a

So the upper limit can be written as a + 6

We get

(a + (a + 6))/ 2 = 10

By cross multiplication

2a + 6 = 20

On further calculation

2a = 14

By division

a = 7

We know that

Class size = 20 – 15 = 5

Consider a as the lower limit of the required class

Upper limit can be written as a + 5

So we get

(a + (a + 5))/ 2 = 20

On further calculation

2a + 5 = 40

So we get

2a = 35

By division

a = 17.5

Upper limit a + 5 = 17.5 + 5 = 22.5

Therefore, the required class is 17.5-22.5

Correct option is (C) Arithmetic Mean

The mid-value of the class is used to calculate arithmetic mean.

C) Arithmetic Mean

We know that lower limit of the lowest class = 10

It is given that

Width of each class = 5

Total number of classes = 5

So we get

First class = 10-15

Second class = 15-20

Third class = 20-25

Fourth class = 25-30

Fifth class = 30-35

Therefore, the upper class limit of the highest class is 35.

Correct option is (B) 14.5

Mid-value of the class 10–19 \(=\frac{10+19}2\)

= 14.5

Correct option is  B) 14.5

(C) 35

Explanation:

According to the question,

Since width of each five consecutive classes = 5

And, the lower limit of lowest class = 10

We get the data as follows,

10-15

15-20

20-25

25-30

30-35

Hence, we get upper limit of highest class = 35

Hence, option (C) is the correct answer.

Correct option is (B) bi-modal data

A data having two modes is called bi-modal data.

B) bi-modal data

Correct option is (A) Maximum Value – Minimum Value

For a data, range = Maximum value - minimum value

A) Maximum Value – Minimum Value

C) P is primary data & Q is secondary data

First five prime numbers 2, 3, 5, 7, 11. 

Maximum number = 11 

Minimum number = 2 

Range = 11 – 2 = 9

n = No. of observations

\(\overline x \) = Mean

Boys median class = \(\frac {37+1}{2} = \frac {38}{2}\) = 19th

observation 

∴ Median height of boys = 147 cm

Girls median class = \(\frac {29+1}{2} = \frac {30}{2}\) = 15th

observation 

∴ Median height of girls = 152 cm

Weight of Ranga = 46 kg 

Weight of Reshma = Weight of Rahim = x kg say

Average = Sum of the weights/ Number = 40 kg

∴ 40 = \(\frac {46+x+x}{3}\)

3 x 40 = 46 + 2x 

2x = 120 – 46 = 74

∴ x = 74/2 = 37

∴ Rahim’s weight = 37 kg.