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(c) 24

Mode = (3 × median) – (2 × mean) 

⇒ (3 × median) = (mode + 2 mean) 

⇒ (3 × median) = 16 + 56

 ⇒ (3 × median) = 72 

⇒ Median = 72/3 

∴ Median = 24

There is an empirical relationship between the three measures of central tendency:

3 median = mode + 2 Mean

Mean = 3 Median - Mode/2

= 3(1250) - 1000/2

= 1375

Thus, the mean is 1375.

(b) 24.5

Mode = (3 × median) – (2 × mean) 

⇒ (2 × mean) = (3 × median) – mode 

⇒ (2 × mean) = 3 × 26 – 29 

⇒ (2 × mean) = 49

⇒ Mean = 49/2 

∴ Mean = 24.5

(c) mean = mode = median 

A symmetric distribution is one where the left and right hand sides of the distribution are roughly equally balanced around the mean.

There is an empirical relationship between the three measures of central tendency: 3median = mode + 2Mean

⇒ mean = \(\cfrac{3median - mode}2\)

= \(\cfrac{3(1250)-1000}2\)

=1375 

Thus, the mean is 1375.

We know that

All prime numbers between 50 and 80 = 53, 59, 61, 67, 71, 73 and 79

So we get

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (53 + 59 + 61 + 67 + 71 + 73 + 79)/7

On further calculation

Mean = 463/7

So we get

Mean = 66 1/7

Therefore, the mean of all prime numbers between 50 and 80 is 66 1/7.

(a) Mean

The mean cannot be determined graphically because the values cannot be summed.

Correct answer is (d) ogives

This because median of a frequency distribution is found graphically with the help of ogives.

Correct answer is (c) a histogram

The mode of a frequency distribution can be obtained graphically from a histogram

We know that

All the factors of 20 = 1, 2, 4, 5, 10 and 20

So we get

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (1 + 2 + 4 + 5 + 10 + 20)/6

On further calculation

Mean = 42/6

By division

Mean = 7

Therefore, the mean of all the factors of 20 is 7.

We know that

First seven multiples of five = 5, 10, 15, 20, 25, 30 and 35

So we get

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (5 + 10 + 15 + 20 + 25 + 30 + 35)/7

On further calculation

Mean = 140/7

By division

Mean = 20

Therefore, the mean of first seven multiples of five is 20.

By arranging the numbers in ascending order

We get

0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6

From the table we know that 6 occurs maximum number of times so the mode is 6.

No. of terms = 5 and mean = 8 

Sum of numbers = 5 x 8 = 40 ..(i) 

but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii) 

from (i) and (ii) 

27 + y + x = 40 

x + y = 13 

y = 13 – x

By arranging the numbers in ascending order

We get

15, 20, 22, 23, 25, 25, 25, 27, 40

From the table we know that 25 occurs maximum number of times so the mode is 25.

It is given that

Mean height of 30 boys = 150cm

So the total height = 150 (30) = 4500cm

We know that

Correct sum = 4500 – incorrect value + correct value

By substituting the values

Correct sum = 4500 – 135 + 165 = 4350

So we get

Correct mean = Correct sum/30

By substituting the values

Correct mean = 4530/30 = 151 cm

Therefore, the correct mean is 151 cm.

It is given that

Mean of six numbers = 23

So we get the sum of six numbers = 23 (6) = 138

It is given that

Mean of five numbers = 20

So we get the sum of five numbers = 20 (5) = 100

We know that

Excluded number = sum of six numbers – sum of five numbers

By substituting the values

Excluded number = 138 – 100 = 38

Therefore, the excluded number is 38.

Consider x₁, x₂, …….. x₂₄ as the given numbers

So we get

Mean = (x₁ + x₂, …….. + x₂₄)/24

It is given that mean = 35

(x₁ + x₂, …….. + x₂₄)/24 = 35

By cross multiplication

x₁ + x₂, …….. + x₂₄ = 840 ……. (1)

Take (x₁ + 3), (x₂ + 3), ……… (x₂₄ + 3) as new numbers

So the mean of new numbers = [(x₁ + 3) + (x₂ + 3) + ……… + (x₂₄ + 3)]/24

From equation (1) we get

[(x₁ + 3) + (x₂ + 3) + ……… + (x₂₄ + 3)]/24 = (840 + 72)/24

On further calculation

[(x₁ + 3) + (x₂ + 3) + ……… + (x₂₄ + 3)]/24 = 912/24

By division

Mean of new numbers = 38

Therefore, the mean of the new numbers is 38.

We know that

First ten odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19

So we get

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10

On further calculation

Mean = 100/10

By division

Mean = 10

Therefore, the mean of first ten odd numbers is 10.

Consider x₁, x₂, ……. x₂₀ as the given numbers

So we get

Mean = (x₁ + x₂ + ……. + x₂₀)/20

It is given that mean = 43

(x₁ + x₂ + ……. + x₂₀)/20 = 43

By cross multiplication

x₁ + x₂ + ……. + x₂₀ = 860 …… (1)

Take (x₁ – 6), (x₂ – 6) …… (x₂₀ – 6) as the new numbers

So the mean of new numbers = [(x₁ – 6) + (x₂ – 6) + …… + (x₂₀ – 6)]/20

From equation (1) we get

[(x₁ – 6) + (x₂ – 6) + …… + (x₂₀ – 6)]/20 = (860 – 120)/20

On further calculation

[(x₁ – 6) + (x₂ – 6) + …… + (x₂₀ – 6)]/20 = 740/20

By division

Mean of new numbers = 37

Therefore, the new mean of numbers is 37.

No. of terms = 5 

Mean = 8 Sum of numbers = 8 x 5 = 40 .   (i) 

But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii) From (i) and (ii) 

27+a = 40 

a = 13

Consider x₁, x₂, ……. x₁₅ as the given numbers

So we get

Mean = (x₁ + x₂ + ……. + x₁₅)/15

It is given that mean = 27

(x₁ + x₂ + ……. + x₁₅)/15 = 27

By cross multiplication

x₁ + x₂ + ……. + x₁₅ = 405 …… (1)

Take (x₁ × 4), (x₂ × 4), ……… (x₁₅ × 4) as new numbers

So the mean of new numbers = [(x₁ × 4) + (x₂ × 4) + ……… + (x₁₅ × 4)]/15

From equation (1) we get

[(x₁ × 4) + (x₂ × 4) + ……… + (x₁₅ × 4)]/15 = (405 × 4)/ 15

On further calculation

Mean of new numbers = 1620/15 = 108

Therefore, the mean of new numbers is 108.

Consider x₁, x₂, ……. x₁₂ as the given numbers

So we get

Mean = [x₁ + x₂ + ……. + x₁₂]/12

It is given that mean = 40

(x₁ + x₂ + ……. + x₁₂)/12 = 40

By cross multiplication

x₁ + x₂ + ……. + x₁₂ = 480 …… (1)

Take (x₁ ÷ 8), (x₂ ÷ 8), ……… (x₁₂ ÷ 8) as new numbers

So the mean of new numbers = [(x₁ ÷ 8) + (x₂ ÷ 8) + ……… + (x₁₂ ÷ 8)]/12

From equation (1) we get

[(x₁ ÷ 8) + (x₂ ÷ 8) + ……… + (x₁₂ ÷ 8)]/12 = (480 ÷ 8)/12

On further calculation

Mean of new numbers = 60/12 = 5

Therefore, the mean of the new numbers is 5.

Consider x₁, x₂, ……. x₂₀ as the given numbers

So we get

Mean = (x₁ + x₂ + ……. + x₂₀)/20

It is given that mean = 18

(x₁ + x₂, ……. + x₂₀)/20 = 18

By cross multiplication

x₁ + x₂ + ……. + x₂₀ = 360 …… (1)

Take (x₁ + 3), (x₂ + 3), ……… (x₁₀ + 3) as first ten new numbers

So the mean of new set of 20 numbers = [(x₁ + 3) + (x₂ + 3) + ……… + (x₁₀ + 3) + x₁₁ + …….. + x₂₀]/ 20

We get

Mean of new set of 20 numbers = [(x₁ + x₂ + ……. + x₂₀) + 3 × 10]/ 20

From equation (1) we get

Mean of new set of 20 numbers = (360 + 30)/ 20 = 19.5

Therefore, the mean of new set of 20 numbers is 19.5.

(c) a histogram

The mode of a frequency distribution can be obtained graphically from a histogram.

(d) ogives

This because median of a frequency distribution is found graphically with the help of ogives.

It is given that

Average marks in 4 subjects = 50

So the total marks = 50 (4) = 200

We know that

36 + 44 + 75 + x = 200

On further calculation

155 + x = 200

So we get

x = 200 – 155

By subtraction

x = 45

Therefore, the value of x is 45.

Arranging the given data in descending order: 8, 7, 6, 5, 4, 3, 3, 1, 0 The middle term is 4 which is the 5th term. Median = 4