Measures of Central Tendency (Mean)

Find the mean of each of the following sets of numbers :

(i) 10, 4, 6, 12, 9

(ii) 0.2, 0.02, 2, 2.02, 1.22, 1.02

Answer

(i) Given,

10, 4, 6, 12, 9

We know that,

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{10 + 4 + 6 + 12 + 9}{5} \\[1em] = \dfrac{41}{5} \\[1em] = 8.2$

Hence, mean of given numbers = 8.2.

(ii) Given,

0.2, 0.02, 2, 2.02, 1.22, 1.02

We know that,

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{0.2 + 0.02 + 2 + 2.02 + 1.22 + 1.02}{6} \\[1em] = \dfrac{6.48}{6} \\[1em] = 1.08.$

Hence, mean of given numbers = 1.08.

Find the arithmetic mean of :

(i) first eight natural numbers;

(ii) first five prime numbers;

(iii) first six positive even integers;

(iv) first five positive integral multiples of 3;

(v) all factors of 20.

Answer

(i) Given,

first eight natural numbers = 1, 2, 3, 4, 5, 6, 7, 8

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \dfrac{36}{8} \\[1em] = 4.5$

Hence, mean of given numbers = 4.5.

(ii) Given,

first five prime numbers = 2, 3, 5, 7, 11

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 3 + 5 + 7 + 11}{5} \\[1em] = \dfrac{28}{5} \\[1em] = 5.6$

Hence, mean of given numbers = 5.6.

(iii) Given,

first six positive even integers; = 2, 4, 6, 8, 10, 12

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 4 + 6 + 8 + 10 + 12}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7$

Hence, mean of given numbers = 7.

(iv) Given,

first five positive integral multiples of 3 = 3, 6, 9, 12, 15

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{3 + 6 + 9 + 12 + 15}{5} \\[1em] = \dfrac{45}{5} \\[1em] = 9.$

Hence, mean of given numbers = 9.

(v) Given,

all factors of 20 = 1, 2, 4, 5, 10, 20

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 4 + 5 + 10 + 20}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7.$

Hence, mean of given numbers = 7.

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

Find the mean temperature of the week.

Answer

We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean of temperatures } = \dfrac{\text{Sum of temperatures of all days}}{\text{Number of days}} \\[1em] = \dfrac{35.5 + 30.8 + 28.3 + 31.1 + 23.8 + 29.9 + 32.7}{7} \\[1em] = \dfrac{212.1}{7} \\[1em] = 30.3 ^{\circ} F$

Hence, mean temperature of the week is 30.3 °F.

The marks obtained by 10 students in a class-test were as follows :

38, 41, 36, 31, 45, 38, 27, 32, 29, 39

Find :

(i) the mean of their marks;

(ii) the mean of their marks, when the marks of each student are increased by 2;

(iii) the mean of their marks, when 1 mark is deducted from the marks of each student;

(iv) the mean of their marks, when the marks of each student are halved.

Answer

(i) We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean of marks} = \dfrac{\text{Sum of marks}}{\text{Number of students}} \\[1em] = \dfrac{38 + 41 + 36 + 31 + 45 + 38 + 27 + 32 + 29 + 39}{10} \\[1em] = \dfrac{356}{10} \\[1em] = 35.6$

Hence, mean of marks = 35.6.

(ii) If 2 marks are added to each student, the mean also increases by 2.

New mean = 35.6 + 2 = 37.6

Hence, mean when marks are increased by 2 = 37.6

(iii) If 1 mark is deducted to each student, the mean also decreases by 1.

New mean = 35.6 - 1 = 34.6

Hence, mean when 1 mark is deducted = 34.6

(iv) If the marks of each student are halved, the mean is also halved.

New mean = $\dfrac{35.6}{2}$ = 17.8

Hence, mean when marks are halved = 17.8

If the mean of 11, 8, 13, 10, x and 9 is 9.5, find the value of x.

Answer

We know that,

Mean = $\dfrac{\sum x_i}{n}$

We have,

$\Rightarrow \text{Mean} = \dfrac{11 + 8 + 13 + 10 + x + 9}{6} \\[1em] \Rightarrow 9.5 = \dfrac{51 + x}{6} \\[1em] \Rightarrow 6(9.5) = 51 + x \\[1em] \Rightarrow 57 = 51 + x \\[1em] \Rightarrow x = 57 - 51 \\[1em] \Rightarrow x = 6.$

Hence, value of x is 6.

Find the mean of 25 numbers, it being given that the mean of 15 of them is 18 and the mean of remaining ones is 13.

Answer

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

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of 10 numbers is 13

∴ Sum of 13 terms = 13 × 10 = 130

Sum of 25 terms = 130 + 270 = 400.

Mean = $\dfrac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 25 of them is 51 kg, find the mean weight of the remaining students.

Answer

By formula,

Mean = $\dfrac{\text{Sum of weight of students}}{\text{No.of students}}$

Given,

Mean weight of 60 students of a class = 52.75 kg

$\therefore 52.75 = \dfrac{\text{Sum of weight of students}}{60}$

Total weight = 60 × 52.75

= 3165 kg.

Mean weight of 25 students among them = 51 kg.

So, the total weight of 25 students = 51 × 25 = 1275 kg.

Remaining students = 60 – 25 = 35

Total weight of remaining 35 students = 3165 – 1275 = 1890 kg

Mean weight of 35 students = $\dfrac{1890}{35}$ = 54 kg.

Hence, the mean weight of the remaining students is 54 kg.

The mean of five numbers is 18. On excluding one number, the mean becomes 16. Find the excluded number.

Answer

Given:

Number of observations = 5

Mean = 18

⇒ Sum of all 5 observations = 5 x 18 = 90

On excluding an observation, the mean of the remaining 4 observations = 16

∵ Sum of all remaining 4 observations = 4 x 16 = 64

⇒ Excluded observation = Sum of all 5 observations - Sum of all remaining 4 observations

= 90 - 64

= 26

Hence, the excluded number is 26.

The ages of 40 students of a group are given below :

Find the mean age of the group.

Answer

We know that,

n = ∑f = 40.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{580}{40}$ = 14.5 years

Hence, mean age of the group is 14.5 years.

Find the mean of the following frequency distribution :

Answer

We know that,

n = ∑f = 50.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{359}{50}$ = 7.18

Hence, mean of the frequency distribution is 7.18.

In a book of 300 pages, the distribution of misprints is shown below :

Find the average number of misprints per page.

Answer

We know that,

n = ∑f = 300.

By formula,

Mean = $\dfrac{\sum fx}{n} = \dfrac{222}{300}$ = 0.74 per page.

Hence, average number of misprints per page is 0.74.

The following table gives the wages of different categories of workers in a factory :

(i) Calculate the mean wage.
(ii) If the number of workers in each category is doubled, what would be the new mean wage?

Answer

(i) We know that,

n = ∑f = 50.

By formula,

Mean wage = $\dfrac{\sum fx}{n} = \dfrac{21200}{50}$ = ₹ 424

Hence, mean wage is ₹ 424.

(ii) If all frequencies in a distribution are multiplied by a constant, the mean remains unchanged.

n = ∑f = 50 x 2 = 100.

∑fx = 21200 x 2 = 42400

By formula,

Mean wage = $\dfrac{\sum fx}{n} = \dfrac{42400}{100}$ = ₹ 424

Hence, new mean wage is ₹ 424.

If the mean of the following distribution is 7.5, find the missing frequency f :

Answer

We know that,

n = ∑f = 74 + f.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 7.5 = \dfrac{563 + 7f}{74 + f}.$

⇒ 7.5(74 + f) = 563 + 7f

⇒ 555 + 7.5f = 563 + 7f

⇒ 7.5f - 7f = 563 - 555

⇒ 0.5f = 8

⇒ f = $\dfrac{8}{0.5}$

⇒ f = 16.

Hence, missing frequency f is 16.

If the mean of the following observations is 16.6, find the numerical value of p.

Answer

We know that,

n = ∑f = 76 + p.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 16.6 = \dfrac{1228 + 18p}{76 + p}.$

⇒ 16.6(76 + p) = 1228 + 18p

⇒ 1261.6 + 16.6p = 1228 + 18p

⇒ 1261.6 - 1228 = 18p - 16.6p

⇒ 33.6 = 1.4p

⇒ p = $\dfrac{33.6}{1.4}$

⇒ p = 24.

Hence, numerical value of p is 24.

Find the numerical value of x, if the mean of the following frequency distribution is 12.58.

Answer

We know that,

n = ∑f = 50.

By formula,

$\text{Mean} = \dfrac{\sum fx}{n} \\[1em] 12.58 = \dfrac{524 + 7x}{50}.$

⇒ 12.58(50) = 524 + 7x

⇒ 629 = 524 + 7x

⇒ 7x = 629 - 524

⇒ 7x = 105

⇒ x = $\dfrac{105}{7}$

⇒ x = 15.

Hence, numerical value of x is 15.

Using short cut method, compute the mean height from the following frequency distribution :

Answer

Let assumed mean (A) = 62.

We know that,

n = ∑f = 80.

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{n} \\[1em] = 62 + \dfrac{28}{80} \\[1em] = 62 + 0.35 \\[1em] = 62.35.$

Hence, mean height of the plants is 62.35 cm.

The number of match sticks contained in 50 match boxes is given below :

(i) Using short cut method, find the mean number of match sticks per box.

(ii) How many extra match sticks are to be added to all the contents of 50 match boxes to bring the mean exactly equal to 45 match sticks per box?

Answer

(i) We know that,

n = ∑f = 50.

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{n} \\[1em] = 44 + \dfrac{-16}{50} \\[1em] = 44 - 0.32 \\[1em] = 43.68$

Hence, mean number of match sticks per box is 43.68

(ii) Total number of sticks = Mean × Total Boxes

= 43.68 × 50

= 2184 sticks

Number of match sticks to be added = (50 × 45) − 2184 = 66.

Hence, extra match sticks to be added = 66.

The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method.

Answer

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{\sum f} \\[1em] = 25 + \dfrac{90}{40} \\[1em] = 25 + 2.25 \\[1em] = 27.25.$

Hence, required mean = 27.25

Given below are the daily wages of 200 workers in a factory :

Calculate the mean daily wages.

Answer

By formula,

$\text{Mean} = A + \dfrac{\sum fd}{\sum f} \\[1em] = 390 + \dfrac{9000}{200} \\[1em] = 390 + 45 \\[1em] = 435.$

Hence, mean daily wage is ₹435.

If the mean of the following distribution is 24, find the value of a.

Answer

By formula,

$\text{Mean} = \dfrac{\sum f_iy_i}{\sum f_i} \\[1em] 24 = \dfrac{810 + 15a}{30 + a} \\[1em] 24(30 + a) = 810 + 15a \\[1em] 720 + 24a = 810 + 15a \\[1em] 24a - 15a = 810 - 720 \\[1em] 9a = 90 \\[1em] a = 10$

Hence, the value of a = 10.

Calculate the mean of the following distribution using step deviation method.

Answer

We construct the following table, taking assumed mean a = 25. Here, c (width of each class) = 10.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 25 + 10 \times \dfrac{63}{100} \\[1em] = 25 + \dfrac{630}{100} \\[1em] = 25 + 6.3 \\[1em] = 31.3$

Hence, mean of the following distribution is 31.3

Using step deviation method, calculate the mean of the following frequency distribution :

Answer

We construct the following table, taking assumed mean a = 85. Here, c (width of each class) = 10.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 85 + 10 \times \dfrac{6}{75} \\[1em] = 85 + \dfrac{4}{5} \\[1em] = 85 + 0.8 \\[1em] = 85.8$

Hence, mean of the given frequency distribution is 85.8.

The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram, by the step Deviation Method.

Answer

We construct the following table, taking assumed mean a = 97.5. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 97.5 + 5 \times \dfrac{-16}{50} \\[1em] = 97.5 - \dfrac{16}{10} \\[1em] = 97.5 - 1.6 \\[1em] = 95.9 \approx 96$

Hence, mean weight of the apples is 96 g.

Weights of 60 eggs were recorded as given below :

Calculate their mean weight to the nearest gm.

Answer

Since, class are discontinuous we will first convert them into continuous class intervals.

Adjustment factor

= $\dfrac{\text{Lower limit of a class -Upper limit of previous class}}{2} = \dfrac{80 - 79}{2} = \dfrac{1}{2}$ = 0.5

Adding the adjustment factor to upper limit and subtracting from lower limit we get the continuous class intervals.

We construct the following table, taking assumed mean a = 92. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 92 + 5 \times \dfrac{-19}{60} \\[1em] = 92 - \dfrac{19}{12} \\[1em] = 92 - 1.583 \\[1em] = 90.416 \approx 90$

Hence, mean weight of the eggs is 90 g.

The following table gives marks scored by students in an examination :

Calculate the mean marks correct to 2 decimal places.

Answer

We construct the following table, taking assumed mean a = 17.5. Here, c (width of each class) = 5.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 17.5 + 5 \times \dfrac{17}{80} \\[1em] = 17.5 + \dfrac{85}{80} \\[1em] = 17.5 + 1.0625 \\[1em] = 18.5625 \approx 18.56$

Hence, mean marks scored by the students is 18.56

The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method.

Take the assumed mean as 45. Give your answer correct to 2 decimal places.

Answer

We construct the following table, taking assumed mean a = 45.

By formula,

$\text{Mean} = a + \dfrac{\sum f_id_i}{\sum f_i} \\[1em] = 45 + \dfrac{-140}{30} \\[1em] = 45 - \dfrac{14}{3} \\[1em] = 45 - 4.6666 \\[1em] = 40.3333 \approx 40.33$

Hence, average number of patients attending the hospital per month is 40.33.

Calculate the mean of the following frequency distribution.

Answer

We construct the following table, taking assumed mean a = 30. Here, c (width of each class) = 10.

By formula,

$\text{Mean} = a + c \times \dfrac{\sum f_iu_i}{\sum f_i} \\[1em] = 30 + 10 \times \dfrac{6}{24} \\[1em] = 30 + \dfrac{10}{4} \\[1em] = 30 + 2.5 \\[1em] = 32.5$

Hence, mean of the frequency distribution is 32.5.

Which of the following is not a measure of central tendency?

1. Mean

2. Mode

3. Range

4. Median

Answer

Range is a measure of dispersion, not central tendency, it describes how spread out the data is.

Range is the difference between the highest and lowest values of data.

Hence, option 3 is the correct option.

The mean of the following data is : 34, 89, 37, 144, 78, 240, 128, 98

1. 102

2. 104

3. 106

4. 108

Answer

By formula,

$\text{Mean} = \dfrac{\sum x_i}{n} \\[1em] = \dfrac{34 + 89 + 37 + 144 + 78 + 240 + 128 + 98}{8} \\[1em] = \dfrac{848}{8} \\[1em] = 106.$

Hence, option 3 is the correct option.

If the mean of 7, 5, 13, x and 9 be 10, then the value of x is :

1. 10

2. 12

3. 14

4. 16

Answer

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum x_i}{n} \\[1em] \Rightarrow 10 = \dfrac{7 + 5 + 13 + x + 9}{5} \\[1em] \Rightarrow 10 \times 5 = 34 + x \\[1em] \Rightarrow 50 = 34 + x \\[1em] \Rightarrow x = 50 - 34 \\[1em] \Rightarrow x = 16.$

Hence, option 4 is the correct option.

In a monthly test, the marks obtained in Mathematics by 16 students of a class are as follows :

0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

The arithmetic mean of the marks obtained is :

1. 3

2. 4

3. 5

4. 6

Answer

By formula,

$\text{Arithmetic mean} = \dfrac{\text{ Sum of all marks}}{\text{ Number of students}} \\[1em] = \dfrac{0 + 0 + 2 + 2 + 3 + 3 + 3 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 7 + 8}{16} \\[1em] = \dfrac{64}{16} \\[1em] = 4.$

Hence, option 2 is the correct option.

Out of 100 numbers, 20 were 4s, 40 were 5s, 30 were 6s and the remaining were 7s. The mean of the numbers is :

1. 5.3

2. 5.4

3. 6.1

4. 6.5

Answer

By formula,

$\text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] = \dfrac{530}{100} \\[1em] = 5.3.$ Hence, option 1 is the correct option.

If 36 a + 36 b = 576, then the mean of a and b is :

1. 6

2. 8

3. 12

4. 16

Answer

Given,

36a + 36b = 576

36(a + b) = 576

a + b = $\dfrac{576}{36}$

a + b = 16

By formula,

$\text{Mean} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{a + b}{2} \\[1em] = \dfrac{16}{2} \\[1em] = 8.$

Hence, option 2 is the correct option.

If the mean of 7 observations is 43 and each observation is increased by 7, then what will be the new mean?

1. 36

2. 43

3. 44

4. 50

Answer

Given,

Mean = 43

When each observation is increased by 7, mean will also increase by 7, thus new mean = 43 + 7 = 50.

Hence, option 4 is the correct option.

While computing the mean of grouped data, we assume that the frequencies are :

1. evenly distributed over all the classes

2. centred at the class marks of the classes

3. centred at the upper limits of the classes

4. centred at the lower limits of the classes

Answer

To calculate the mean, we need a single representative value for each class. We assume that the data points in that interval are balanced around the middle, known as the Class Mark.

Class mark = $\dfrac{\text{Upper limit + lowerlimit}}{2}$

Hence, option 2 is the correct option.

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the mean of first three observations is :

1. 9

2. 11

3. 13

4. none of these

Answer

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum x_i}{n} \\[1em] \Rightarrow 11 = \dfrac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} \\[1em] \Rightarrow 11 \times 5 = 5x + 20 \\[1em] \Rightarrow 55 = 5x + 20 \\[1em] \Rightarrow 55 - 20 = 5x \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \dfrac{35}{5} \\[1em] \Rightarrow x = 7.$

Mean of first three observations, x = 7, x + 2 = 7 + 2 = 9, x + 4 = 7 + 4 = 11

$\text{Mean} = \dfrac{\sum x_i}{n} \\[1em] = \dfrac{7 + 9 + 11}{3} \\[1em] = \dfrac{27}{3} \\[1em] =9.$

Hence, option 1 is the correct option.

The mean of five consecutive odd numbers A, B, C, D and E in ascending order is 37. What is the product of B and D?

1. 1365

2. 1585

3. 1935

4. 2035

Answer

Given,

Five consecutive odd numbers :

A = x - 4,

B = x - 2,

C = x,

D = x + 2,

E = x + 4.

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum x_i}{n} \\[1em] \Rightarrow 37 = \dfrac{(x - 4) + (x - 2) + x + (x + 2) + (x + 4)}{5} \\[1em] \Rightarrow 37 = \dfrac{5x}{5} \\[1em] \Rightarrow x = 37.$

C = x = 37

B = 37 - 2 = 35

D = 37 + 2 = 39

B × D = 35 × 39 = 1365.

Hence, option 1 is the correct option.

The mean of 5 consecutive odd numbers of set A is 37. What will be the average of set B containing four consecutive even numbers if the smallest number of set B is 13 more than the greatest number of set A?

1. 53

2. 55

3. 57

4. 59

Answer

Five consecutive odd numbers = x - 4, x - 2, x, x + 2, x + 4.

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum x_i}{n} \\[1em] \Rightarrow 37 = \dfrac{(x - 4) + (x - 2) + x + (x + 2) + (x + 4)}{5} \\[1em] \Rightarrow 37 \times 5 = 5x \\[1em] \Rightarrow x = 37.$

Set A = 33, 35, 37, 39, 41

Smallest number of Set B = 41 + 13 = 54

Set B = 54, 56, 58, 60

$\text{Average} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{54 + 56 + 58 + 60}{4} \\[1em] = \dfrac{228}{4} \\[1em] = 57.$

Hence, option 3 is the correct option.

If the mean of four observations is 20 and when a constant c is added to each observation the mean becomes 22. The value of c is :

1. −2

2. 2

3. 4

4. 6

Answer

Given,

The mean of four observations is 20

When a constant c is added to each observation, total increase is 4(c)

New sum = 80 + 4c

New mean = $\dfrac{80 + 4c}{4} = 22$ 80 + 4c = 22(4)

4c = 88 - 80

c = $\dfrac{8}{4}$

c = 2.

Hence, option 2 is the correct option.

If the mean of the observations x₁, x₂, x₃, ……, x_n is x̄, then the mean of x₁ − a, x₂ − a, x₃ − a, ……, x_n − a is :

1. $\bar x$

2. $\bar x$ + a

3. $\bar x$ − a

4. $\Big(\dfrac{n \bar x - a}{n}\Big)$

Answer

The original mean =

$\bar{x} = \dfrac{x_1 + x_2 + x_3 + \dots + x_n}{n}$

The new mean is the sum of the new observations divided by :

$\text{New Mean} = \dfrac{(x_1 - a) + (x_2 - a) + (x_3 - a) + \dots + (x_n - a)}{n} \\[1em] = \dfrac{(x_1 + x_2 + \dots + x_n) - (a + a + \dots + a)}{n}$

Since there are n terms of a, the sum of the a is na:

$\text{New Mean} = \dfrac{(x_1 + x_2 + \dots + x_n) - na}{n}\\[1em] = \dfrac{x_1 + x_2 + \dots + x_n}{n} - \dfrac{na}{n} \\[1em] = \bar{x} - a.$

Hence, option 3 is the correct option.

The mean of a certain number of observations is x̄. If each observation is multiplied by m (m ≠ 0) and then increased by n, then the mean of new observations is :

1. $\Big(\dfrac{\bar x}{m} + n\Big)$

2. m $\bar x$ + n

3. m $\bar x$ − n

4. $\Big(\dfrac{\bar x}{m} - n\Big)$

Answer

Let the original observations be x₁, x₂, ......., x_k with a mean of $\bar{x}$.

The original mean is:

$\bar{x} = \dfrac{\sum x_i}{k}$

If every observation is multiplied by m, the new observations are, mx₁, mx₂,......., mx_k.

The sum of these new values is

$\sum mx_i = m(\sum x_i)$.

Adding "n" to every term increases mean by "n":

$\text{Mean} = \dfrac{m(\sum x_i)}{k} = m \bar{x} + n$

Hence, option 2 is the correct option.

The mean of 100 observations is 50. If one of the observations was misread as 50 instead of 40, the correct mean is :

1. 40

2. 49.9

3. 50

4. 50.1

Answer

Sum of observations when observations were misread = Number of observations × Initial Mean

= 100(50)

= 5000

To find the correct sum We need to subtract the wrong value and add the correct value,

Sum = 5000 - 50 + 40

= 4990

∴ Mean = $\dfrac{4990}{100}$ = 49.9

Hence, option 2 is the correct option.

If the mean of the following data is 25, the value of p is equal to :

1. 2

2. 3

3. 4

4. 5

Answer

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] \Rightarrow 25 = \dfrac{390 + 15p}{14 + p} \\[1em] \Rightarrow 25(14 + p) = 390 + 15p \\[1em] 350 + 25p = 390 + 15p \\[1em] \Rightarrow 25p - 15p = 390 - 350 \\[1em] 10p = 40 \\[1em] \Rightarrow p = \dfrac{40}{10} \\[1em] \Rightarrow p = 4.$

Hence, option 3 is the correct option.

Consider the table given below :

The mean of the marks given above is :

1. 6

2. 18

3. 27

4. 28

Answer

By formula,

$\text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] = \dfrac{2800}{100} \\[1em] = 28.$

Hence, option 4 is the correct option.

If the mean of the following distribution is 27, then the value of p is :

1. 6

2. 7

3. 9

4. 11

Answer

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] \Rightarrow 27 = \dfrac{1245 + 15p}{43 + p} \\[1em] \Rightarrow 27(43 + p) = 1245 + 15p \\[1em] 1161 + 27p = 1245 + 15p \\[1em] \Rightarrow 27p - 15p = 1245 - 1161 \\[1em] 12p = 84 \\[1em] \Rightarrow p = \dfrac{84}{12} \\[1em] \Rightarrow p = 7.$

Hence, option 2 is the correct option.

The mean of 5 numbers is 27. If one of the numbers be excluded, their mean is 25. The excluded number is :

1. 25

2. 26

3. 28

4. 35

Answer

Given,

Mean of 5 numbers = 27

Sum = Number of observations × Initial Mean

= 5(27)

= 135

When one number is excluded, 4 numbers remain, and their new mean is 25.

Sum of numbers(when one number is excluded) = Number of observations × new mean

= 4(25)

= 100

The difference between the two sums is the value of the number that was removed = 135 - 100 = 35

Hence, option 4 is the correct option.

For what value of x the mean of the given observations (2x − 5), (x + 3), (7 − x), (5 − x) and (x + 9) with frequencies 2, 3, 4, 6 and 1 respectively is 4?

1. 1

2. 2

3. 3

4. 4

Answer

By formula,

$\text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] 4 = \dfrac{66 - 2x}{16} \\[1em] 2x = 66 - 64 \\[1em] 2x = 2 \\[1em] x = 1.$

Hence, option 1 is the correct option.

The mean of n observations is $\bar x$. If the first observation is increased by 1, second by 2 and so on, then the new mean is :

1. $\bar x$ + n

2. $\bar x$ + $\Big(\dfrac{n}{2}\Big)$

3. $\bar x$ + $\Big(\dfrac{n + 1}{2}\Big)$

4. $\bar x$ + $\Big(\dfrac{n − 1}{2}\Big)$

Answer

Let the n observations x₁, x₂, ......., x_n. The mean is $\bar x$.

$\bar{x} = \dfrac{x_1 + x_2 + ... + x_n}{n} \\[1em] \sum x = n\bar x$

Given,

The first observation is increased by 1, second by 2 and so on.

New sum of observations = (x₁ + 1)+ (x₂ + 2)+ .......+ (x_n + n)

= (x₁+ x₂+ .......+ x_n) + (1 + 2 + ..... + n)

= ∑x + (1 + 2 + 3 + ... + n)

$= n\bar x + \dfrac{n(n + 1)}{2}$

$\text{ Mean }= \dfrac{\text{ Sum of observations}}{\text{ Number of observations}} \\[1em] = \dfrac{n\bar x + \dfrac{n(n + 1)}{2}}{n} \\[1em] = \bar x + \dfrac{n + 1}{2}.$

Hence, option 3 is the correct option.

In the formula $\bar x = a + \Big(\dfrac{\sum f_i d_i}{\sum f_i}\Big)$ for finding the mean of grouped data, d_is are deviations from a, of :

1. lower limits of the classes

2. upper limits of the classes

3. mid-points of the classes

4. frequencies of the class marks

Answer

Given,

$\bar x = a + \Big(\dfrac{\sum f_i d_i}{\sum f_i}\Big)$

the deviation d_i are calculated as:

d_i = x_i - a

where x_i are the class marks.

So, the deviations are taken from mid-points of the classes.

Hence, option 3 is the correct option.

If x_i's are the class marks of the class-intervals of grouped data, f_i's are the corresponding frequencies and x̄ is the mean, then Σ f_i(x_i − x̄) is equal to :

1. −1

2. 0

3. 1

4. 2

Answer

Given,

$\Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \sum f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \bar x \sum f_i \text{ .....(1)}$

We know that,

$\bar x = \dfrac{\sum f_i x_i}{\sum f_i} \\[1em] \bar x \sum f_i = \sum f_i x_i$

Substituting the values in equation (1), we get :

$\therefore \sum f_i(x_i - \bar x) =\bar x \sum f_i - \bar x \sum f_i = 0$

Hence, option 2 is the correct option.

In the formula x̄ = a + h $\Big(\dfrac{\Sigma f_i u_i}{\Sigma f_i}\Big)$ for finding the mean of grouped frequency distribution, u_i =

1. $\Big(\dfrac{x_i + a}{h}\Big)$

2. h (x_i − a)

3. $\Big(\dfrac{x_i − a}{h}\Big)$

4. $\Big(\dfrac{a − x_i}{h}\Big)$

Answer

In the Step-Deviation Method formula $\bar x = a + h \Big(\dfrac{\sum f_i u_i}{\sum f_i}\Big)$,

the term u_i represents the step-deviation.

The step-deviation is calculated as:

$u_i = \frac{x_i - a}{h}$

Where :

x_i : The class mark (mid-point).

a : The assumed mean.

h : The class size.

Hence, option 3 is the correct option.

In a class of 100 students, the mean marks obtained in a certain test is 30 and in another class of 50 students the mean marks obtained in the same test is 60. The mean marks obtained by the students of both the classes taken together is :

1. 40

2. 45

3. 48

4. 50

Answer

Sum of marks obtained in class 1 = 100(30) = 3000

Sum of marks obtained in class 2 = 50(60) = 3000

Total marks obtained in class 1 and class 2 = 6000

Total number of students in class 1 and class 2 = 100 + 50 = 150

By formula,

$\text{Mean} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{6000}{150} \\[1em] = 40.$

Hence, option 1 is the correct option.

A distribution consists of three components with frequencies 45, 40 and 15 having their means 2, 2.5 and 2 respectively. The mean of the combined distribution is :

1. 2.1

2. 2.2

3. 2.3

4. 2.4

Answer

$\text{Mean} = \dfrac{\sum f_ix_i}{\sum f_i} \\[1em] = \dfrac{220}{100} \\[1em] = 2.2$

Hence, option 2 is the correct option.

The combined mean of three groups is 12 and the combined mean of first two groups is 3. If the first, second and third groups have 2, 3 and 5 items respectively, then the mean of third group is :

1. 10

2. 12

3. 13

4. 21

Answer

Total number of items in three groups = 2 + 3 + 5 = 10

Total sum = Number of observations × Mean

= 10 × 12 = 120

Total number of items in first two groups = 2 + 3 = 5

Sum of first two groups = Number of observations × Mean

= 3 × 5 = 15

The sum of the third group is the difference between the total sum and the sum of the first two groups = 120 - 15 = 105.

By formula,

$\text{Mean} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{105}{5} \\[1em] = 21.$

Hence, option 4 is the correct option.

The mean weight of 17 boxes is 92 kg. If 18 new boxes are added, the mean weight increases by 3 kg. What will be the mean weight of the 18 new boxes?

1. 91.8 kg

2. 92.8 kg

3. 97.8 kg

4. 98.8 kg

Answer

Total weight of 17 boxes = Mean × Number of boxes

= 92 × 17 = 1564 kg.

If 18 new boxes are added, mean increases by 3 kg.

So, new mean = 92 + 3 = 95 kg.

Total weight of 35 boxes = New mean × Number of boxes

= 95 × 35

= 3325 kg

The weight added by the 18 new boxes = Total weight of 35 boxes - Total weight of 17 boxes

= 3325 - 1564

= 1761 kg.

By formula,

$\text{Mean of 18 boxes} = \dfrac{\text{Weight of 18 boxes}}{\text{ Number of boxes}} \\[1em] = \dfrac{1761}{18} \\[1em] = 97.8 \text{ kg.}$

Hence, option 3 is the correct option.

Consider the following distribution :

If the mean of the above distribution is 50, what is the value of f ?

1. 24

2. 34

3. 56

4. 96

Answer

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum f_i x_i}{\sum f_i} \\[1em] \Rightarrow 50 = \dfrac{4320 + 70f}{96 + f} \\[1em] \Rightarrow 50(96 + f) = 4320 + 70f \\[1em] 4800 + 50f = 4320 + 70f \\[1em] \Rightarrow 4800 - 4320 = 70f - 50f \\[1em] \Rightarrow 480 = 20f \\[1em] \Rightarrow f = \frac{480}{20} \\[1em] \Rightarrow f = 24.$

Hence, option 1 is the correct option.

Which of the following cannot be determined graphically?

1. Mean

2. Median

3. Mode

4. none of these

Answer

The mean is an algebraic measure that cannot be determined graphically.

To calculate mean, we must sum all observations and divide by the total number. Since it depends on the exact value of every single observation rather than their position or frequency alone there is no standard geometric construction or curve that can pinpoint the mean on a graph.

Hence, option 1 is the correct option.

Directions : The marks obtained by 10 students in a class-test were as follows :

36, 64, 48, 52, 57, 73, 26, 39, 78, 67

31. The mean marks of the whole class is :

(a) 49.1
(b) 53.7
(c) 54
(d) 60

32. If the maximum marks in the test were 80, the mean percentage of marks obtained by the students is :

(a) 65%
(b) 67.5%
(c) 68%
(d) 72%

33. The mean marks of the top 5 scorers in the class is :

(a) 65.4
(b) 66.8
(c) 67.2
(d) 67.8

34. As per the Board’s instruction each student who obtained less than 50 marks was awarded 3 grace marks. The new mean of the marks thus obtained increases by :

(a) 0.9
(b) 1.2
(c) 1.5
(d) 1.8

Answer

31. By formula,

$\text{Mean} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{36 + 64 + 48 + 52 + 57 + 73 + 26 + 39 + 78 + 67}{10} \\[1em] = \dfrac{540}{10} \\[1em] = 54.$

Hence, option (c) is the correct option.

32. By formula,

$\text{Percentage of marks} = \dfrac{\text{Mean of marks obtained}}{\text{Maximum marks}} \times 100 \\[1em] = \dfrac{54}{80} \times 100 \\[1em] = 0.675 \times 100 \\[1em] = 67.5$

Hence, option (b) is the correct option.

33. Top five marks in class are 78, 73, 67, 64, 57.

By formula,

$\text{Mean} = \dfrac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \dfrac{78 + 73 + 67 + 64 + 57}{5} \\[1em] = \dfrac{339}{5} \\[1em] = 67.8$

Hence, option (d) is the correct option.

34. The students who scored less than 50 are: 36, 48, 26, 39

Since each of these 4 students gets 3 marks,

Total marks increased = 4(3) = 12 marks

The new mean of the marks thus obtained increases by :

$\dfrac{\text{Total marks increased}}{\text{Number of students}} \\[1em] = \dfrac{12}{10} \\[1em] = 1.2$

Hence, option (b) is the correct option.

Directions :
At a courier company, a daily report of parcels received for dispatch is prepared every evening, which classifies the parcels on the basis of their weights. The report of a certain day is as under :

35.How many parcels have weights in the range of 300 – 400 grams?
(a) 4
(b) 12
(c) 13
(d) 19

36.How many parcels have weights in the range of 200 – 300 grams?
(a) 10
(b) 12
(c) 13
(d) 19

37.In which of the following weight ranges, the number of parcels is the lowest?
(a) 100 – 200
(b) 200 – 300
(c) 400 – 500
(d) 500 – 600

38.The mean weight of parcels received on that particular day is :
(a) 226.78 g
(b) 251.67 g
(c) 284.28 g
(d) 302.16 g

Answer

Table :

35. From table,

The number of parcels for the 300 – 400 range = 19 (54 - 35).

Hence, option (d) is the correct option.

36.From table,

The number of parcels for the 200 – 300 range = 13 (35 - 22).

Hence, option (c) is the correct option.

37.The lowest frequency is 2, which occurs in the 500 – 600 range.

Hence, option (d) is the correct option.

38. By formula,

$\text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] = \dfrac{15100}{60} \\[1em] = 251.67 \text{ g.}$

Hence, option (b) is the correct option.

Assertion (A): For a grouped frequency distribution, we use Mean = A + $\Big(\dfrac{\Sigma f t}{\Sigma f}\Big)$ × h to find the mean using step deviation method.

Reason (R): Here t = $\Big(\dfrac{x − A}{h}\Big)$.

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

The standard formula to calculate the mean ($\bar{x}$) using the step-deviation method is:

$\bar{x} = A + \Big( \dfrac{\sum ft}{\sum f} \Big) \times h$

Therefore, Assertion (A) is true.

The step-deviation is defined as the difference between the class mark (x) and the assumed mean (A), divided by the class size (h):

$t = \dfrac{x - A}{h}$

Reason (R) is true.

Hence, option 3 is the correct option.

Assertion (A): If x_i’s are the mid-points of the class intervals of a grouped data, Σf_i’s are the corresponding frequencies and x̄ is the mean, then Σf_i(x_i − x̄) = 1.

Reason (R): The sum of the deviations from the mean is 0.

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

The expression $\sum f_i(x_i - \bar{x})$ represents the sum of the deviations of all observations from their mean, weighed by their frequencies.

Given,

$\Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \sum f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \bar x \sum f_i \text{ .....(1)}$

We know that,

$\bar x = \dfrac{\sum f_i x_i}{\sum f_i} \\[1em] \bar x \sum f_i = \sum f_i x_i$

Substituting the values in equation 1,

$\therefore \sum f_i(x_i - \bar x) =\bar x \sum f_i - \bar x \sum f_i = 0$

Assertion (A) is false.

The sum of the deviations from the mean is 0. This is a fundamental and correct property of the arithmetic mean.

Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): Out of 25 numbers, the mean of 15 of them is 18. If the mean of the remaining numbers is 13, then the mean of the 25 numbers is 14.

Reason (R): Mean of the variates x₁, x₂, …, x_n having corresponding frequencies f₁, f₂, …, f_n is given by

x̄ = $\Big(\dfrac{f_1 x_1 + f_2 x_2 + \dots + f_n x_n}{x_1 + x_2 + \dots + x_n}\Big)$.

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

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

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of remaining numbers is 13

∴ Sum of remaining terms = 13 × 10 = 130

Sum of 25 terms = 130 + 270 = 400.

Mean = $\dfrac{400}{25} = 16$

Assertion (A) is false.

The correct formula is,

$\bar x = \dfrac{\sum f_ix_i}{\sum f_i}$

Reason (R) is false.

Hence, option 4 is the correct option.