Mean & Median of Ungrouped Data & Frequency Polygon

The weight of 7 boys in a group are 52 kg, 57 kg, 55 kg, 60 kg, 54 kg, 59 kg and 55 kg. Find the mean weight of the group.

Answer

Weights of 7 boys (in kg) :

52, 57, 55, 60, 54, 59, 55

Number of boys (n) = 7

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

Sum of observations = 52 + 57 + 55 + 60 + 54 + 59 + 55 = 392

Total number of observations = 7

Mean = $\dfrac{392}{7}$ = 56.

Hence, mean weight of the group is 56 kg.

The marks obtained by 7 students in a group are 340, 180, 260, 164, 56, 275 and 307 respectively. Find the mean marks per student.

Answer

Marks of 7 students:

340, 180, 260, 164, 56, 275, 307

Number of students (n) = 7

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

Sum of observations = 340 + 180 + 260 + 164 + 56 + 275 + 307 = 1582

Total number of observations = 7

Mean = $\dfrac{ 1582}{7}$ = 226.

Hence, mean marks per student is 226.

Find the mean of first six prime numbers.

Answer

First six prime numbers are:

2, 3, 5, 7, 11, 13

Mean = $\dfrac{\text{Sum of prime numbers}}{\text{Total number of prime numbers}}$

Sum of prime numbers = 2 + 3 + 5 + 7 + 11 + 13 = 41

Mean = $\dfrac{41}{6} \approx$ 6.83

Hence, mean of first six prime numbers is 6.83.

Find the mean of the first ten odd numbers.

Answer

First ten odd numbers are:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Sum of observations = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{100}{10}$ = 10.

Hence, mean of first ten odd numbers is 10.

Find the mean of all the factors of 20.

Answer

Factors of 20 are the numbers that divide 20 exactly.

The factors are :

1, 2, 4, 5, 10, 20

Total number of observations = 6

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{1 + 2 + 4 + 5 + 10 + 20}{6}$

= $\dfrac{42}{6}$ = 7.

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

The daily minimum temperature recorded (in degree F) at a place during a week was as under:

Find the mean temperature.

Answer

We know that,

Mean = $\dfrac{\text{Sum of temperatures of all days}}{\text{Number of days}}$

= $\dfrac{35.2 + 31.1 + 27.6 + 31.8 + 29.3 + 23.8}{6}$

= $\dfrac{178.8}{6}$

= 29.8 °F

Hence, mean temperature = 29.8 °F.

If the mean of 6, 8, 9, x, 13 is 10, find the value of x.

Answer

Numbers: 6, 8, 9, x, 13

Mean = 10

Number of observations (n) = 5

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

⇒ 10 = $\dfrac{6 + 8 + 9 + x + 13}{5}$

⇒ 50 = 36 + x

⇒ x = 50 - 36 = 14.

Hence, x = 14.

The mean of the heights of 6 girls is 148 cm. If the individual heights of five of them are 142 cm, 154 cm, 146 cm, 145 cm and 150 cm, find the height of the sixth girl.

Answer

Mean height of 6 girls = 148 cm

Number of girls (n) = 6

Heights of 5 girls:

142 cm, 154 cm, 146 cm, 145 cm, 150 cm

Let height of the sixth girl be x cm.

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

⇒ 148 = $\dfrac{142 + 154 + 146 + 145 + 150 + x}{6}$

⇒ 148 × 6 = 737 + x

⇒ 888 = 737 + x

⇒ x = 888 - 737 = 151.

Hence, height of the sixth girl = 151 cm.

The following table shows the weights (in kg) of 15 workers in a factory :

Calculate the mean weight.

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (weight)

So, ∑fx = 975 and ∑f = 15

Therefore,

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

= $\dfrac{975}{15}$ = 65.

Hence, mean weight = 65 kg.

Find the mean of daily wages of 60 workers in a factory as per data given below:

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (Daily wages)

so, ∑fx = 7110 and ∑f = 60

Mean = $\dfrac{7110}{60}$ = 118.5.

Hence, mean of daily wages of 60 workers is ₹118.5

The heights (in cm) of 90 plants in a garden are given below :

Calculate the mean height.

Answer

$\text{Mean} = \dfrac{\sum fx}{\sum f}$

where

f = frequency (Number of plants)

x = observation (Height)

∑fx = 5634 and ∑f = 90

Mean = $\dfrac{5634}{90}$ = 62.6

Hence, mean height = 62.6 cm.

The mean of the following data is 21.6. Find the value of p.

Answer

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

∑f_i = 5 + 4 + p + 6 + 4 + 6 = 25 + p

∑f_ix_i​ = 30 + 48 + 18p + 144 + 120 + 216 = 558 + 18p

⇒ 21.6 = $\dfrac{558+18p}{25+p}$

⇒ 21.6(25 + p) = 558 + 18p

⇒ 540 + 21.6p = 558 + 18p

⇒ 21.6p − 18p = 558 − 540

⇒ 3.6p = 18

⇒ p = $\dfrac{18}{3.6}$ = 5

Hence, p = 5.

If the mean of the following data is 18.75, find the value of p.

Answer

Mean = 18.75

∑f_i​ = 5 + 10 + 7 + 8 + 2 = 32

∑f_ix_i​ = 50 + 150 + 7p + 200 + 60 = 460 + 7p

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

⇒ 18.75 = $\dfrac{460 + 7p}{32}$

⇒ 600 = 460 + 7p

⇒ 7p = 600 - 460

⇒ 7p = 140

⇒ p = $\dfrac{140}{7}$ = 20.

Hence, p = 20.

The mean age of a group of 40 students is 17.45 years. Find the missing frequencies.

Answer

Let missing frequencies be:

Frequency at age 16 = a

Frequency at age 19 = b

By formula,

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

where

x_i = Age (in years)

f_i = Number of students

∑f_i = 3 + a + 9 + 11 + b + 3 = 40

⇒ a + b + 26 = 40

⇒ a + b = 14

⇒ b = 14 - a ........(1)

Mean = $\dfrac{\sum f_ix_i}{\sum f_i}$

⇒ 17.45 = $\dfrac{45 + 16a + 153 + 198 + 19b + 60}{40}$

⇒ 17.45 × 40 = 456 + 16a + 19b

⇒ 698 = 456 + 16a + 19b

⇒ 242 = 16a + 19b ..........(2)

Substitute equation 1 in equation 2

⇒ 16a + 19(14 − a) = 242

⇒ 16a + 266 − 19a = 242

⇒ −3a + 266 = 242

⇒ −3a = −24

⇒ a = 8.

⇒ Put a = 8 in equation 1

⇒ b = 14 - 8

⇒ b = 6.

Hence, the missing frequencies are 8 & 6.

Using the assumed mean method, calculate the mean weekly wage from the following frequency distribution :

Answer

By using assumed mean method,

Let A = 1100

By formula,

Mean = A + $\dfrac{\sum f_id_i}{\sum f_i}$

= 1100 + $\dfrac{5850}{110}$

= 1100 + 53.18

= 1153.18.

Hence, mean weekly wage is ₹ 1153.18.

Using the step-deviation method, find the mean from the following data :

Answer

We construct the following table, taking assumed mean A = 21

Here c (width of the class) = 1

Mean = A + c × $\dfrac{\sum f_iu_i}{\sum f_i}$

= 21 + 1 × $\dfrac{-840}{2200}$

= 21 + $\dfrac{-840}{2200}$

= 21 - 0.382 = 20.62.

Hence, mean of following distribution is 20.62.

Find the median of :

(i) 15, 6, 16, 8, 22, 21, 9, 18, 25

(ii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27

(iii) 55, 60, 35 ,51, 29, 63, 72, 91, 85, 82

Answer

(i) By arranging data in ascending order, we get :

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

Number of observations, n = 9, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ $\dfrac{9 + 1}{2}$ th observation

⇒ $\dfrac{10}{2}$th observation

⇒ 5th observation

∴ Median = 16.

Hence, median is 16.

(ii) By arranging data in ascending order, we get :

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81

Number of observations, n = 12, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{12}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{12}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{6th term} + \text{7th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{15} + \text{17}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{32}}{2} \\[1em]$

∴ Median = 16.

Hence, median = 16.

(iii) By arranging data in ascending order, we get :

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of observations, n = 10, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2}\\[1em] \Rightarrow \text{Median} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{60} + \text{63}}{2}\\[1em] \Rightarrow \text{Median} = \dfrac{\text{123}}{2}$

∴ Median = 61.5.

Hence, median is 61.5.

The runs scored by 11 members of a cricket team are :

26, 38, 53, 18, 66, 72, 0, 47, 32, 7, 35

Find the median score.

Answer

By arranging data in ascending order, we get :

0, 7, 18, 26, 32, 35, 38, 47, 53, 66, 72

Number of observations, n = 11, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ $\dfrac{11 + 1}{2}$ th observation

⇒ $\dfrac{12}{2}$th observation

⇒ 6th observation

∴ Median = 35.

Hence, median is 35 runs.

The heights (in cm) of 9 girls are :

144.2, 148.5, 152.1, 143.7, 145, 149.6, 150, 146.5, 147.3

Find the median height.

Answer

By arranging data in ascending order, we get :

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Number of observations, n = 9, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ $\dfrac{9 + 1}{2}$ th observation

⇒ $\dfrac{10}{2}$th observation

⇒ 5th observation

∴ Median = 147.3.

Hence, median is 147.3 cm.

The age(in years) of 10 teachers in a school are :

34, 37, 53, 46, 52, 43, 31, 36, 40, 50

Find the median age.

Answer

By arranging data in ascending order, we get :

31, 34, 36, 37, 40, 43, 46, 50, 52, 53

Number of observations, n = 10, which is even.

By formula,

$\text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow\text{Median} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{40} + \text{43}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{83}}{2} \\[1em]$

∴ Median = 41.5.

Hence, median is 41.5 years.

The weights (in kg) of 8 children are :

13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8

Find the median weight.

Answer

By arranging data in ascending order, we get :

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Number of observations, n = 8, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{8}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{8}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{4th term} + \text{5th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{13.4} + \text{14.3}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{27.7}}{2} \\[1em]$

∴ Median = 13.85.

Hence, median is 13.85 kg.

Find the median weight for the following data:

Answer

Cumulative frequency table :

Total number of observations = 41, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ $\dfrac{41 + 1}{2}$ th observation

⇒ $\dfrac{42}{2}$th observation

⇒ 21st observation

From the table,

From 20th to 28th position, the weight = 50 kg

∴ Median = 50 kg.

Hence, median weight is 50 kg.

Calculate the median for the following frequency distribution :

Answer

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Total number of observations = 40, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{40}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{40}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{20th term} + \text{21st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{7} + \text{10}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{17}}{2} \\[1em]$

∴ Median = 8.5.

Hence, median is 8.5.

The heart of 60 patients were examined through X-ray and the observations obtained are given below :

Find the median.

Answer

Cumulative frequency table :

Total number of observations = 60, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{60}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{122} + \text{122}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{244}}{2} \\[1em]$

∴ Median = 122.

Hence, median = 122 mm.

Find the median for the following data :

Answer

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Total number of observations = 60, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{60}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{20} + \text{23}}{2} \\[1em] \Rightarrow\text{Median} = \dfrac{\text{43}}{2} \\[1em]$

∴ Median = 21.5.

Hence, median is 21.5.

The following table shows the marks obtained by the students of a class in an examination.

Draw a frequency polygon.

Answer

Frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 10 marks.

2. Take 2 cm along y-axis = 10 students.

3. Find the mid-point of class intervals.

4. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

5. Join consecutive points by line segments.

6. Join first end point with mid-point of class (-10) - 0 with zero frequency and join the other end with mid-point of class 50 - 60 with zero frequency.

The required frequency polygon is shown below:

Draw a frequency polygon to represent the following data:

Answer

Frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 9 units.

2. Take 2 cm along y-axis = 5 units.

3. Find the mid-point of class intervals.

4. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

5. Join consecutive points by line segments.

6. Join first end point with mid-point of class (-9) - 0 with zero frequency and join the other end with mid-point of class 45 - 54 with zero frequency.

The required frequency polygon is shown below:

The heights of boys in a school are given below :

Draw a frequency polygon to represent the above data.

Answer

Frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 5 cm.

2. Take 2 cm along y-axis = 10 boys.

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 140.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 135 - 140 with zero frequency and join the other end with mid-point of class 165 - 170 with zero frequency.

The required frequency polygon is shown below :

Draw a frequency polygon to represent the following data :

Answer

Frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 5 kg.

2. Take 2 cm along y-axis = 5 workers.

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 35.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 30 - 35 with zero frequency and join the other end with mid-point of class 60 - 65 with zero frequency.

The required frequency polygon is shown below :

Draw a frequency polygon to represent the following data:

Answer

Frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = ₹ 100.

2. Take 1 cm along y-axis = 5 workers.

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 750.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 650 - 750 with zero frequency and join the other end with mid-point of class 1250 - 1350 with zero frequency.

The required frequency polygon is shown below:

Construct a frequency polygon from the following data :

Answer

The following frequency distribution table is discontinuous. Convert it into continuous frequency distribution.

Adjustment factor = $\dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2}$

= $\dfrac{6 - 5}{2} = \dfrac{1}{2}$ = 0.5.

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution table :

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 5 units.

2. Take 2 cm along y-axis = 2 units.

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 0.5.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class -5.5 - 0.5 with zero frequency and join the other end with mid-point of class 25.5 - 30.5 with zero frequency.

The required frequency polygon is shown below:

The mean of the numbers : 7, 9, 4, 6, 5 is :

1. 5.8

2. 6.0

3. 6.2

4. 6.4

Answer

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{7 + 9 + 4 + 6 + 5}{5}$

= $\dfrac{31}{5}$ = 6.2

Hence, option 3 is the correct option.

The median of the data 2, 7, 9, 13, 20, 22, 24, 25, 27, 28, 35, 40, is :

1. 24

2. 23

3. 25

4. 27

Answer

The data is already arranged in the ascending order.

Number of observations, n = 12, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{12}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{12}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{6th term} + \text{7th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{22} + \text{24}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{46}}{2} \\[1em]$

∴ Median = 23.

Hence, option 2 is the correct option.

Following data have been arranged in the ascending order.

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.

If the median of the data is 63, the value of x is :

1. 31

2. 62

3. 124

4. 134

Answer

Here n = 10, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{63} = \dfrac{\text{x} + \text{x+2}}{2} \\[1em]$

⇒ 126 = 2x + 2

⇒ 2x = 124

⇒ x = 62.

Hence, option 2 is the correct option.

If the mean of 10, 12, 18, 13, 20 and 17 is :

1. 14

2. 15

3. 16

4. 18

Answer

By arranging data in ascending order we get :

10, 12, 13, 17, 18, 20

Number of observations (n) = 6

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{10+12+13+17+18+20}{6}$

= $\dfrac{90}{6}$ = 15.

Hence, option 2 is the correct option.

If the mean of first 8 prime numbers is :

1. 9.625

2. 8.625

3. 10.625

4. 12

Answer

First 8 prime numbers are :

2, 3, 5, 7, 11, 13, 17, 19

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

= $\dfrac{2+3+5+7+11+13+17+19}{8}$

= $\dfrac{77}{8}$ = 9.625.

Hence, option 1 is the correct option.

If the mean of 10, 12, 16, 20, p and 26 is 17, then the value of p is :

1. 16

2. 18

3. 20

4. 24

Answer

Mean = 17

Number of observations = 6

Mean = $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$

⇒ 17 = $\dfrac{10+12+16+20+p+26}{6}$

⇒ 102 = 84 + p

⇒ p = 102 - 84

⇒ p = 18.

Hence, option 2 is the correct option.

Mean of 10 observations is 20 and that of other 15 observations is 16. Mean of all 25 observations will be :

1. 16.6

2. 18.6

3. 19.6

4. 17.6

Answer

Mean of 10 observations = 20

Mean of 15 observations = 16

Mean = $\dfrac{\text{Total sum}}{\text{Total number of observations}}$

Sum of 10 observations = 10 × 20 = 200

Sum of 15 observations = 15 × 16 = 240

Total Sum = 200 + 240 = 440

Total number of observations = 25

Mean = $\dfrac{440}{25}$ = 17.6.

Hence, option 4 is the correct option.

The height of 8 students in a class are 102 cm, 107 cm, 82 cm, 103 cm, 97 cm, 111 cm, 88 cm, and 94 cm. The mean height is :

1. 94 cm

2. 95 cm

3. 97 cm

4. 98 cm

Answer

Mean = $\dfrac{\text{Sum of all heights}}{\text{Total number of students}}$

= $\dfrac{102+107+82+103+97+111+88+94}{8}$

= $\dfrac{784}{8}$ = 98 cm.

Hence, option 4 is the correct option.

If the mean of the data x₁, x₂, x₃, ....., x_n is 'a', then the mean of the data x₁ + a, x₂ + a, ...., x_n + a is :

1. a

2. 2a

3. $\dfrac{1}{2}$a

4. $\dfrac{a}{3}$

Answer

Mean of x₁, x₂, x₃, ....., x_n is 'a'

So,

$\dfrac{x_1 + x_2 + \dots + x_n}{n} = a$

x₁ + x₂ + ..... + x_n = na

∴ $\sum x_i = na$

New observation is :

x₁ + a, x₂ + a, ...., x_n + a

$\Rightarrow \text{New Mean} = \dfrac{(x_1 + a) + (x_2 + a) + \dots + (x_n + a)}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{(x_1 + x_2 + \dots + x_n) + (a + a + \dots + n \text{ times})}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{\sum x_i + na}{n}$

Since, $\sum x_i = na$

∴ $\text{New Mean} = \dfrac{na + na}{n} = \dfrac{2na}{n}$ = 2a.

Hence, option 2 is the correct option.

If the mean of the data y₁, y₂, y₃, ...., y_n is 102, then the mean of the data 5y₁, 5y₂, 5y₃, ...., 5y_n is :

1. 102

2. 204

3. 510

4. 606

Answer

Mean of y₁, y₂, y₃, ...., y_n is 102

So,

$\dfrac{y_1 + y_2 + \dots + y_n}{n} = 102$

New observation is :

5y₁, 5y₂, 5y₃, ...., 5y_n

$\text{New Mean} = \dfrac{5y_1 + 5y_2 + \dots + 5y_n}{n} \\[1em] = \dfrac{5(y_1 + y_2 + \dots + y_n)}{n} \\[1em] = 5 \times \left( \dfrac{y_1 + y_2 + \dots + y_n}{n} \right)$

Since,

$\dfrac{y_1 + y_2 + \dots + y_n}{n} = 102$

∴ $\text{New Mean} = 5 \times 102$ = 510.

Hence, option 3 is the correct option.

The average score of Raman in last 5 innings is 42. If he scored 38, 41, 43 and 47 in first 4 innings, then his score in the fifth innings was :

1. 31

2. 41

3. 51

4. 61

Answer

Average score of 5 innings = 42

⇒ Total score = 42 × 5 = 210

Total score of first 4 innings = 38 + 41 + 43 + 47 = 169

Fifth innings score = 210 - 169 = 41.

Hence, option 2 is the correct option.

The mean of 25 observations was calculated to be 44. Later on it was found that two of the observations 34 and 46 were wrongly copied as 28 and 42. The correct mean will be :

1. 39.4

2. 40.4

3. 44.4

4. 43.9

Answer

Given

Wrong mean = 44

Total observations = 25

So,

Total wrong sum = 25 × 44 = 1100

Two numbers were wrongly copied:

Wrong numbers → 28 and 42

Correct numbers → 34 and 46

⇒ 34 - 28 = 6

⇒ 46 - 42 = 4

⇒ Increased sum = 6 + 4 = 10

So, the correct sum = 1100 + 10 = 1110

∴ Correct mean = $\dfrac{1110}{25}$ = 44.4

Hence, option 3 is the correct option.

The median of the following observations : 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. The value of x is :

1. 19

2. 20

3. 21

4. 21.5

Answer

Given,

Median = 24

Set of numbers arranged in ascending order,

11, 12, 14, 18, (x + 4), 30, 32, 35, 41

Number of observations (n) = 9, which is odd.

By formula,

Median = $\dfrac{n + 1}{2}$ th observation

⇒ 24 = $\dfrac{9 + 1}{2}$ th observation

⇒ 24 = $\dfrac{10}{2}$th observation

⇒ 24 = 5th observation

⇒ 24 = x + 4

⇒ x = 24 - 4

⇒ x = 20.

Hence, option 2 is the correct option.

If the mean of x - 3, x - 1, 7, x, 2x - 1 and 3x - 5 is 3.5, then the median is :

1. 2.5

2. 3.5

3. 3.8

4. 3.9

Answer

Given,

Mean = 3.5

Total number of observations = 6

Mean = $\dfrac{\text{Sum of Observations}}{\text{Total number of Observations}}$

⇒ 3.5 = $\dfrac{x - 3 + x - 1 + 7 + x + 2x - 1 + 3x - 5}{6}$

⇒ 3.5 = $\dfrac{8x - 3}{6}$

⇒ 21 = 8x - 3

⇒ 8x = 24

⇒ x = 3.

∴ The observations are :

0, 2, 7, 3, 5, 4

By arranging the data in ascending order, we get :

0, 2, 3, 4, 5, 7

Number of observations (n) = 6, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{6}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{6}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{3rd term} + \text{4th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{3} + \text{4}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{7}{2} \\[1em]$

∴ Median = 3.5.

Hence, option 2 is the correct option.

The runs scored by a cricketer in last 20 innings are given below.

The mean runs of the cricketer per inning is :

1. 21

2. 24

3. 29

4. 30

Answer

Total innings = 20

Arranging the table into Runs and frequency :

0 × 1 = 0

5 × 3 = 15

17 × 6 = 102

32 × 5 = 160

61 × 3 = 183

70 × 2 = 140

⇒ Total runs = 0 + 15 + 102 + 160 + 183 + 140 = 600

⇒ Mean = $\dfrac{\text{Total Runs}}{\text{Total innings}}$

= $\dfrac{600}{20}$ = 30.

Hence, option 4 is the correct option.

Case Study:
A school organized a Health Check Up Camp for its students. The weights (in kg) of the students of a class were recorded as below :

41, 40, 36, 52, 50, 48, 47, 45,
40, 41, 42, 49, 50, 51, 38, 41,
40, 45, 40, 39, 49, 51, 48, 46,
44, 50, 57, 38, 41, 51

Based on the above information, answer the following questions:

1. The range of the data is :
   (a) 20 kg
   (b) 21 kg
   (c) 22 kg
   (d) 23 kg

2. Mean weight of the data is :
   (a) 45 kg
   (b) 44.5 kg
   (c) 42.5 kg
   (d) 40.5 kg

3. Median of the data is :
   (a) 44 kg
   (b) 45 kg
   (c) 45.5 kg
   (d) 46 kg

4. One student weighing 45 kg, was absent on that day. Next day, the teacher added his name in the list. The average weight of the new group is :
   (a) 47 kg
   (b) 46.4 kg
   (c) 46 kg
   (d) 45 kg

5. The median weight of the students after adding the absentee in the list is :
   (a) 45 kg
   (b) 46 kg
   (c) 44.5 kg
   (d) 44.2 kg

Answer

1. Range = Highest value - Lowest value

= 57 - 36 = 21 kg.

Hence, option (b) is the correct option.

2. Mean = $\dfrac{\text{Total Weight}}{\text{Total number of observations}}$

= $\dfrac{1350}{30}$ = 45 kg.

Hence, option (a) is the correct option.

3. By arranging data in the ascending order, we get :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 30, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{30}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{30}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{15th term} + \text{16th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{45} + \text{45}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{90}{2} \\[1em]$

∴ Median = 45 kg.

Hence, option (b) is the correct option.

4. Earlier total sum = 1350

After adding the new student (weighing 45 kg) total sum = 1350 + 45 = 1395

∴ New mean = $\dfrac{1395}{31}$ = 45 kg.

Hence, option (d) is the correct option.

5. After adding the absentee ( weighing 45 kg) the observations are :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 31, which is odd.

Median = $\dfrac{n + 1}{2}$ th observation

= $\dfrac{31 + 1}{2}$ th observation

= $\dfrac{32}{2}$th observation

= 16th observation

Median = 45 kg.

Hence, option (a) is the correct option.

Assertion (A) : The mean of 19 numbers is 38. If the mean of the first 10 numbers is 36 and that of the last 10 is 40, then the 10th number is 38.

Reason (R) : Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

Mean of 19 number = 38

⇒ Total sum = 19 × 38 = 722

Mean of first 10 numbers = 36

⇒ Total sum = 10 × 36 = 360

Mean of last 10 numbers = 40

⇒ Total sum = 10 × 40 = 400

The 10th number is included in both groups:

⇒ First 10 numbers → includes 10th

⇒ Last 10 numbers → also includes 10th

So,

⇒ 10th number = (Sum of first 10 numbers + Sum of last 10 numbers)- Sum of all the 19 numbers

= (360 + 400) - 722

= 760 - 722 = 38.

∴ Assertion (A) is true.

We know that,

Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

∴ Reason (R) is true.

Both Assertion (A) and Reason (R) are true.

Hence, option 3 is the correct option.

Assertion (A) : The mean of 15 observations was found to be 21. Later it was detected that one value 15 was wrongly copied as 18, while calculating the mean. The correct mean is 20.

Reason (R) : The mean of n observations x₁, x₂, x₃, ....., x_n is $\bar{x}$. If each observation is increased by p, then the new mean is increased by p, i.e., the new mean is $\bar{x}$ + p.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

Wrong mean = 21

Total observations = 15

So,

Total wrong sum = 21 × 15 = 315

One value 15 was wrongly copied as 18

∴ 18 - 15 = 3

New total sum = 315 - 3 = 312

New mean = $\dfrac{312}{15}$

= 20.8.

But given mean = 20.

∴ Assertion (A) is false.

If given observation is increased by p, then

$\Rightarrow \text{New Mean} = \dfrac{(x_1 + p) + (x_2 + p) + \dots + (x_n + p)}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{(x_1 + x_2 + \dots + x_n) + np}{n} \\[1em] \Rightarrow \text{New Mean} = \bar{x} + p \\[1em]$

∴ Reason (R) is true.

Assertion (A) is false, Reason (R) is true.

Hence option 2 is the correct option.

Which of the following statements is/are correct?

(i) Mean of a data can be smaller than the smallest number of the data.

(ii) Mean of a data can be bigger than the largest number of the data.

1. Only (i)

2. Only (ii)

3. Both (i) and (ii)

4. Both are wrong

Answer

(i) If we have a set of numbers, the average cannot be lower than the lowest value because there is no data pulling the average down past that point.

∴ Statement (i) is wrong.

(ii) Mean of a data cannot be bigger than the larger number because the average represents a "middle" value. So, their average cannot mathematically go beyond that value.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

Average marks obtained by students of a class in an exam is 70. Which of the following is definitely incorrect?

(i) One of the students scored more than 100 marks.

(ii) One of the students scored less than 10 marks.

1. Only (i)

2. Only (ii)

3. Both (i) and (ii)

4. None of these

Answer

Average Mean marks = 70

Total marks = 70 × Number of students

(i) This can be possible.

E.g : Suppose 3 students scored :

120, 50, 40

Mean = $\dfrac{210}{3}$ = 70.

So, scoring more than 100 is possible.

∴ Statement (i) is wrong.

(ii) Scoring less than 10 marks is also possible.

E.g : Suppose a student scored :

0, 110, 100

Mean = $\dfrac{210}{3}$ = 70.

So, scoring less than 10 marks is also possible.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

A government agency collected the data of individual incomes of 10,625 people and decided that the individual that have less income than the median of the data are economically poor. Assuming that all the individuals have different incomes, which of the following is correct?

1. The number of economically poor people is less than the number of rich people.

2. The number of economically poor people is more than the number of rich people.

3. Exactly half of the people are economically poor.

4. We can't say anything about the number of economically poor people.

Answer

Total people = 10,625

Since, number of observations is odd

∴ Median = $\dfrac{n + 1}{2}$

= $\dfrac{10625 + 1}{2}$

= $\dfrac{10626}{2}$ = 5313.

So median is the 5313th person.

Number of economically poor people are = 5313 - 1 = 5312.

This means that number of rich people are = 10625 - 5312 = 5313.

∴ Number of economically poor people is less than the number of rich people.

Hence, option 1 is the correct option.

If the mean of a and $\dfrac{1}{a}$ is x, then the mean of a³ and $\dfrac{1}{a^3}$ is :

1. x(4x² - 3)

2. x(x²- 2)

3. x² + 3

4. x²

Answer

Mean of a and $\dfrac{1}{a}$ is x

$\Rightarrow \dfrac{a + \dfrac{1}{a}}{2} = x$

$\Rightarrow a + \dfrac{1}{a} = 2x$

We know that,

(a + b)³ = a³ + b³ + 3ab(a + b)

Substituting $b = \dfrac{1}{a}$ :

$\Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3(a)\left(\dfrac{1}{a}\right)\left(a + \dfrac{1}{a}\right) \\[1em] \Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3\left(a + \dfrac{1}{a}\right) \\[1em]$

Substituting $a + \dfrac{1}{a} = 2x$ into the identity:

$\Rightarrow (2x)^3 = a^3 + \dfrac{1}{a^3} + 3(2x) \\[1em] \Rightarrow 8x^3 = a^3 + \dfrac{1}{a^3} + 6x \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 8x^3 - 6x \\[1em]$

The mean of a³ and $\dfrac{1}{a^3}$ is:

$\Rightarrow \text{Mean} = \dfrac{a^3 + \dfrac{1}{a^3}}{2} \\[1em] \Rightarrow \text{Mean} = \dfrac{8x^3 - 6x}{2} \\[1em] \Rightarrow \text{Mean} = 4x^3 - 3x \\[1em] \Rightarrow \text{Mean} = x(4x^2 - 3)$

Hence, option 1 is the correct option.

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is :

1. 49.5

2. 53

3. 46.5

4. 56.5

Answer

Let the original numbers be :

x₁, x₂, x₃, ....., x₅₀

And their mean be 'a'

So,

$\Rightarrow a = \dfrac{x_1 + x_2 + \dots + x_{50}}{50} \\[1em] \Rightarrow x_1 + x_2 + \dots + x_{50} = 50a \\[1em]$

Each number is subtracted from 53.

∴ The new numbers are :

$(53 - x_1), (53 - x_2), (53 - x_3), \dots, (53 - x_{50})$

Sum of new numbers is :

⇒ $(53 - x_1) + (53 - x_2) + (53 - x_3) + \dots + (53 - x_{50})$

⇒ 53 + 53 + ..... + 53 - (x₁ + x₂ + x₃ + ..... + x₅₀)

⇒ 50 × 53 - 50a

⇒ 2650 - 50a

Mean = $\dfrac{\text{Total Sum}}{\text{Total observations}}$

⇒ -3.5 = $\dfrac{2650 - 50a}{50}$

⇒ -175 = 2650 - 50a

⇒ 50a = 2650 + 175

⇒ a = $\dfrac{2825}{50}$

⇒ a = 56.5

∴ Mean of original numbers = 56.5.

Hence, option 4 is the correct option.

There are 10 observations x₁, x₂, x₃, ....., x₁₀ in a data. The first five elements x₁, x₂, ....., x₅ are placed by x₁ + 5, x₂ + 5, ....., x₅ + 5 respectively. Also, the next five elements x₆, x₇, ....., x₁₀ are placed by x₆ - 5, x₇ - 5, ..., x₁₀ - 5 respectively. Calculate the overall change in the mean of the data.

Answer

Total observations = 10

First 5 numbers increased by 5

∴ Total increase in sum = 5 × 5 = 25

Next 5 numbers decreased by 5

∴ Total decrease in sum = 5 × 5 = 25

So, net change = 25 - 25 = 0

So new total sum = original total sum.

Hence, there is no change in the mean of a data.

In a collection of data, there are 5 numbers and the numbers in the data are 3, 4, 12 and 13. If the median is 12, then what can be maximum value of mean of the data?

Answer

Total numbers = 5

For 5 numbers :

Median = $\dfrac{n + 1}{2}$ observation

⇒ 3rd observation.

For median to be 12

The arrangement should be :

3, 4, 12, 13, x or 3, 4, 12, x, 13.

In order to maximum mean, the value of x should be greatest, thus arrangement will be :

3, 4, 12, 13, x

Since, one of the values must is repeated, so value of x will be equal to 13.

3, 4, 12, 13, 13

Mean = $\dfrac{3 + 4 + 12 + 13 + 13}{5} = \dfrac{45}{5}$ = 9.

Hence, maximum mean value = 9.

In an office, mean of salaries of employees is ₹1,00,000 and median is ₹80,000. An employee will be considered, Grade A officer if his income is at least ₹85,000. Based on the above information check the validity of the following statements.

(i) Number of Grade A officers is more than or equal to number of non Grade A officers.

(ii) Mean of salaries of Grade A officers is more than ₹1,00,000.

Answer

Mean = ₹1,00,000

Median (mid value) = ₹80,000

This means:

50% employees earn ≤ ₹80,000

50% employees earn ≥ ₹80,000

(i) To be a Grade A officer, an employee must earn at least ₹85,000.

Since the income (₹85,000) is higher than the median (₹80,000), So fewer than 50% of the employees can qualify for Grade A.

∴ There will be more non Grade A officers than the Grade A officers.

Hence, statement (i) is false.

(ii) Given data shows that Mean (₹1,00,000) is significantly higher than the Median (₹80,000). This means there is a small group of very high earners pulling the average up.

The overall mean (₹1,00,000) includes non Grade A officers who earn less than ₹85,000. Therefore there group mean will definitely be below ₹85,000.

To pull this anchor of the non-Grade A officers up from less than ₹85,000 to a total average of ₹1,00,000, the Grade A officers must have a mean significantly higher than ₹1,00,000.

∴ Mean salaries of Grade A officers is more than ₹1,00,000.

Hence, statement (ii) is true.