3 Marks Question

Question 13 Marks

The mean weight per student in a group of 7 students is 55kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Answer

The mean weight per student in a group of 7 students = 55kg.

Weight of 6 students (in kg) = 52, 54, 55, 53, 56 and 54

Let the weight of seventh student = x kg.

$\therefore\text{Mean Weight}=\frac{\text{Sum of weights}}{\text{Total no. of students}}$

$\Rightarrow55=\frac{52+54+55+53+56+54+\text{x}}{7}$

$\Rightarrow385=324+\text{x}$

$\Rightarrow\text{x}=385-324$

$\Rightarrow\text{x}=61\text{kg}$

$\therefore$ weight of seventh stuent = 61kg.

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Question 23 Marks

The following observation s have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.

Answer

Total number of observation in the given data is 10 (even number). So median of this data will be mean of $\frac{10}{2}$ i.e, 5^th observation and $\frac{10}{2}+1$ i.e, 6^th observation. So,

$\text{Median of data}=\frac{5^\text{th}\text{observation}+6^\text{th}\text{observation}}{2}$

$\Rightarrow63=\frac{\text{x}+\text{x}+2}{2}$

$\Rightarrow63=\frac{\text{2x}+2}{2}$

$\Rightarrow63=\text{x}+1$

$\Rightarrow\text{x}=62$

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Question 33 Marks

The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.

Answer

Given that the arithmetic mean and mode of a data are 24 and 12 respectively. That is,

MEAN = 24

MODE = 12

We have to find median

We know that

⇒ MODE = 3 × MEDIAN - 2 × MEAN

⇒ 12 = 3 × MEDIAN = 12 + (2 × 24)

⇒ 3 × MEDIAN = 12 + 48

⇒ 3 × MEDIAN = 60

$\Rightarrow\text{MEDIAN}=\frac{60}{3}$

⇒ MEDIAN = 20

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Question 43 Marks

Numbers 50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8 are written in descending order and their median is 25, find x.

Answer

Given the number of observation, n = 8

$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$

$=\frac{\text{2x}+10+\text{2x}-8}{2}$

$=\text{2x}+1$

Given Median = 25

$\therefore\text{2x}+1=25$

$\Rightarrow\text{2x}=24$

$\Rightarrow\text{x}=12$

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Question 53 Marks

If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.

Answer

The given data is 3, 4, 3, 5, 4, 6, 6, x.
The mode is the value which occur maximum number of times, that is, the mode has maximum frequency. If the maximum frequency occurs for more than 1 value, then the number of mode is more than 1 and is not unique.
Here it is given that the mode is 4. So, x must be 4, otherwise it contradicts that the mode is 4. Hence x = 4

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Question 63 Marks

If the median of the scores 1, 2, x, 4, 5 (where 1 < 2 < x < 4 < 5) is 3, then find the mean of the scores.

Answer

The given data is 1, 2, x, 4 and 5. Since 1 < 2 < x < 4 < 5, the given data is already in ascending order.

Here, the number of observation n = 5, which is an odd number.

Hence, the median is:

$\Big(\frac{\text{n}+1}{2}\Big)^\text{th}\text{observation}$

$=\Big(\frac{\text{5}+1}{2}\Big)^\text{th}\text{observation}$

$=3^\text{rd}\text{observation}$

$=\text{x}$

Here, it is given that the median is 3. Hence, we have x = 3.

The mean is:

$\frac{1+2+3+4+5}{5}$

$=\frac{15}{5}$

$=3$

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Question 73 Marks

If the median of scores $\frac{\text{x}}2{},\frac{\text{x}}3{},\frac{\text{x}}4{},\frac{\text{x}}5{}$ and $\frac{\text{x}}6{}$ (where x > 0) is 6, then find the value of $\frac{\text{x}}6{}.$

Answer

Given that the median of the scores $\frac{\text{x}}2{},\frac{\text{x}}3{},\frac{\text{x}}4{},\frac{\text{x}}5{},\frac{\text{x}}{6},$ where x > 0 is 6. The number of scores n is 5, which is an odd number. We have to find $\frac{\text{x}}{6}$

Note that the scores are in descending order. Hence the median is:

$\Big(\frac{\text{n}+1}{2}\Big)^\text{th}\text{score}$

$=\Big(\frac{\text{5}+1}{2}\Big)^\text{th}\text{score}$

$=3^\text{rd}\text{score}$

$=\frac{\text{x}}{4}$

But, it is given that the median is 6. Hence, we have:

$\frac{\text{x}}{4}=6$

$\Rightarrow\text{x}=6\times4$

$\Rightarrow\frac{\text{x}}{6}=4$

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Question 83 Marks

If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.

Answer

The given data is 2, 4, 6, 8, x, y. They are 6 in numbers.

The mean is:

$\frac{2+4+6+8+\text{x}+\text{y}}{6}$

$=\frac{20+\text{x}+\text{y}}{6}$

But, it is given that the mean is 5. Hence, we have

$\frac{20+\text{x}+\text{y}}{6}=5$

$\Rightarrow20+\text{x}+\text{y}=30$

$\Rightarrow\text{x}+\text{y}=30-20$

$\Rightarrow\text{x}+\text{y}=10$

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Question 93 Marks

If $\overline{\text{X}}$ is the mean of the ten natural numbers x₁, x₂, x₃, ..., x₁₀ show that:

$\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)=0$

Answer

We have,

$\overline{\text{X}}=\frac{\text{x}_1+\text{x}_2+\dots+\text{x}_{10}}{10}$

$\Rightarrow\text{x}_1+\text{x}_2+\dots+\text{x}_{10}=10\bar{\text{x}}\dots(1)$

Now, $\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)$

$=(\text{x}_1+\text{x}_2+\dots+\text{x}_{10})-(\bar{\text{x}}+\bar{\text{x}}+\bar{\text{x}}+\text{upto 10 terms})$

$=10\bar{\text{x}}-10\bar{\text{x}}$ [By equation (i)]

$\therefore\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)=0...$

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Question 103 Marks

If M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆, Prove that
(x₁ - M) + (x₂ - M) + (x₃ - M) + (x₄ - M) + (x₅ - M) + (x₆ - M) = 0.

Answer

Let M be the mean of x₁, x₂, x₃, x₄, x₅ and x₆

Then,

$\text{M}=\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\text{x}_6}{6}$

= x₁ + x₂ + x₃+ x₄ + x₅+ x₆ = 6M

To Prove: (x₁ - M) + (x₂ - M) + (x₃ - M) + (x₄ - M) + (x₅ - M) + (x₆ - M) = 0.

Proof: L.H.S.

= (x₁ - M) + (x₂ - M) + (x₃ - M) + (x₄ - M) + (x₅ - M) + (x₆ - M)

= (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) - (M + M + M + M + M + M)

= 6M - 6M

= 0

= R.H.S.

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Question 113 Marks

Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Answer

Values 3, 4, 6, 7, 8, 14

$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$

$\therefore\text{Mean}=\frac{3+4+6+7+8+14}{6}$

$\therefore\text{Mean}=\frac{42}{6}$

$=7$

$\therefore$ Sum of deviation of values from their mean

= (3 - 7) + (4 - 7) + (6 - 7) + (7 - 7) + (8 - 7) + (14 - 7)

= - 4 - 3 - 1 + 0 + 1 + 7

= - 8 + 8 = 0

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Question 123 Marks

Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

Answer

Given the numbers are 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33

Arranging the numbers in ascending order

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

n = 11 (odd)

$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$

$=\Big(\frac{\text{11}+1}{2}\Big)^{\text{th}}\text{value}$

$=6^{\text{th}}\text{value}=58$

If 92 is replaced by 99 and 41 by 43

Then the new values are: 33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

n = 11 (odd)

$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$

$=\Big(\frac{\text{11}+1}{2}\Big)^{\text{th}}\text{value}$

$=6^{\text{th}}\text{value}=58$

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Question 133 Marks

Find the median of the data:
92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Answer

Numbers are 92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Arranging the numbers in ascending order

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

n = 10(even)

$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$

$=\frac{67+69}{2}$

$=\frac{136}{2}=68$

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Question 143 Marks

Find the median of the data:
83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Answer

Numbers are 83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Arranging the numbers in ascending order

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

n = 10 (even)

$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$

$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$

$=\frac{45+54}{2}$

$=\frac{99}{2}=49.5$

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Question 153 Marks

Find the median of the data:
41, 43, 127, 99, 71, 92, 71, 58, 57

Answer

Numbers are 41, 43, 127, 99, 71, 92, 71, 58, 57

Arranging the numbers in ascending order

41, 43, 57, 58, 71, 71, 92, 99, 127

n = 9 (odd)

$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$

$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{value}$

$=5^\text{th}\text{value}=71$

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Question 163 Marks

Find the median of the data:
31, 38, 27, 28, 36, 25, 35, 40

Answer

Numbers are 31, 38, 27, 28, 36, 25, 35, 40

Arranging the numbers in ascending order

25, 27, 28, 31, 35, 36, 38, 40

n = 8 (even)

$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$

$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$

$=\frac{31+35}{2}$

$=\frac{66}{2}=33$

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Question 173 Marks

Find the median of the data:
25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Answer

Numbers are 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Arranging the numbers in ascending order

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

n = 10 (even)

$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$

$=\frac{26+29}{2}$

$=\frac{55}{2}=27.5$

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Question 183 Marks

Find the median of the data:
15, 6, 16, 8, 22, 21, 9, 18, 25

Answer

Numbers are 15, 6, 16, 8, 22, 21, 9, 18, 25

Arranging the numbers in ascending order

6, 8, 9, 15, 16, 21, 22, 25

n = 9 (odd)

$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$

$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{value}$

$=5^\text{th}\text{value}=16$

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Question 193 Marks

Find the median of the data:
133, 73, 89, 108, 94, 104, 94, 85, 100, 120

Answer

Numbers are 133, 73, 89, 108, 94, 104, 94, 85, 100, 120

Arranging the numbers in ascending order

73, 85, 89, 94, 94, 100, 104, 108, 120, 133

n = 10 (even)

$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$

$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$

$=\frac{94+100}{2}$

$=\frac{194}{2}=97$

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Question 203 Marks

Find the median of the data:
12, 17, 3, 14, 5, 8, 7, 15

Answer

Numbers are 12, 17, 3, 14, 5, 8, 7, 15

Arranging the numbers in ascending order

3, 5, 7, 8, 12, 14, 15, 17

n = 8 (even)

$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$

$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$

$=\frac{8+12}{2}$

$=\frac{20}{2}=10$

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