Frequency Distribution Tables And Graphs

Given, observations 10000, 10250, 10790, 9865, 15350, 10110

Clearly, No of observations, n = 6

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, mean of above data

Hence, mean is Rs. 110060.83

Given, observations 10.25, 9, 4.75, 8, 2.65, 12, 2.35

Clearly, No of observations, n = 7

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, mean of above data

Hence, mean is 7

Given, Mean of 8 observations is 25.

Clearly, No of observations, n = 8 and mean,

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

⇒ sum of terms is 200.

If one observation 11 is excluded,

No of terms = 7

Sum of terms = 200 - 11 = 189

Hence, new mean is 27.

Given, mean,

No of observations, n = 9

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

⇒ sum of observations = x₁ + x₂ + …+ x₉ = 342

As, 27 is mistaken for 72, therefore, actual sum of observations will increase by (72 - 27) = 45

⇒ Actual sum of observations = 342 + 45 = 387

Hence, new mean is 43.

Let the no of members in family be 'n'.

Mean age = 25

Let x₁, x₂, …, x_n be ages of family members.

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, 5 years ago

⇒ mean age,

⇒ x₁ + x₂ + …+ x_n = 25n [1]

In present, age of each family member will be increase by 5, therefore ages will be

x₁ + 5, x₂ + 5, …, x_n + 5

⇒ Present mean age,

[From 1]

Hence, Present mean age is 30.

no of members = 40

Mean age two years ago = 11

Let x₁, x₂, …, x₄₀ be ages of family members.

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, two years ago

⇒ mean age,

⇒ x₁ + x₂ + …+ x₄₀ = 440 [1]

In present, age of each family member will be increase by 2, therefore ages will be

x₁ + 2, x₂ + 2, …, x₄₀ + 2

But as a person left the group, let his present age be 'x' years

In this case, Sum of ages will be decrease by 'x' and No of persons will be 39.

⇒ Present mean age,

⇒ 468 = 440 + 80 - x [From 1]

⇒ x = 520 - 468

⇒ x = 52

Present age of that person is 52 years.

Given, observations 5, 8, 10, 15, 22

Clearly, No of observations, n = 5

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, mean of above data

Hence, mean is 12

Now, we know deviation

d_i = x_i - a

in this case, a = 12

d₁ = x₁ - a = 5 - 12 = -7

d₂ = x₂ - a = 8 - 12 = -4

d₃ = x₃ - a = 10 - 12 = -2

d₄ = x₄ - a = 15 - 12 = 3

d₅ = x₅ - a = 22 - 12 = 10

Sum of deviations = (-7) + (-4) + (-2) + 3 + 10 = 0

Hence, sum of deviations from their mean is 0.

Given, No of deviations = 20

Sum of deviations = 100

We know,

Mean deviation

⇒ Mean deviation

⇒ Mean deviation = 5

Hence, mean deviation is 5.

Given, observations are 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14

Let assumed mean, a = 17

We know, mean by assumed mean method is

Where, a is assumed mean and n is no of observations and x_i's are observations.

Therefore, mean for given data

Now, let assumed mean a = 14

Yes, we get same result in both cases !

This shows that, actual mean doesn't depend on assumed mean !

Let Karishma got 'a' marks.

And assumed mean be 'a'.

We know,

Arithmetic mean,

Where, a is assumed mean.

As, deviations given are -8, -6, -3, -1, 0, 2, 3, 4, 6

Sum of deviations = -8 + (-6) + (-3) + (-1) + 0 + 2 + 3 + 4 + 6

= -3

As, arithmetic mean of marks is 15

Putting values in formula, We have,

⇒ 15 = a - 0.3

⇒ a = 15.3

⇒ Karishma got 15.3 marks

Given, no of observations = n

And sum of deviation from 25 is 25 and sum of deviation from 35 is -25.

Let the actual mean be x,

We know that

Arithmetic mean,

Where, a is assumed mean.

In first case, where assumed mean, a = 25 and sum of deviation is 25, Putting values in formula

We have,

[1]

In first case, where assumed mean, a = 35 and sum of deviation is -25, Putting values in formula

We have,

[2]

Adding [1] and [2], we get

⇒ 2x = 60

⇒ x = 30

⇒ arithmetic mean of required data is 30.

For finding median, first we arrange data either in ascending or descending order,

3.1, 3.2, 3.3, 3.5, 3.7, 3.8

No of terms (n) = 6

As, n is even

Median is arithmetic mean of term and term.

⇒ Median is arithmetic mean of 3^rd and 4^th term.

⇒ Median

⇒ Median = 3.4

Given, median = 15

As the terms are in ascending order, and

No of terms, n = 7

As, n is odd

Median is term.

⇒ Median is 4^th term

⇒ median = x - 3

⇒ 15 = x - 3

⇒ x = 18

Hence, value of x is 18.

As, mode is the most frequently occurring value.

Clearly, 10 is most frequent

Therefore, mode is 10.

As, mode is the most frequently occurring value.

'x' is most frequent score, now if each score is decreased by 3.

Each 'x' in scores will also be decreased by 3.

And mode will be (x - 3).

From 1 to 100,

As, mode is the most frequent occurring value.

Mode of above data is 1, as 1 occurs 21 times.

Given, data is 5, 28, 15, 10, 15, 8, 24

No of terms,

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

Therefore, mean of above data

For finding median, let's arrange data in increasing order

5, 8, 10, 15, 15, 24, 28

As, n is even

Median is term.

⇒ Median is 4^th term

⇒ median = 15

Also, mode is the most frequently occurring value.

Clearly, 15 occurs two times and is most frequent

Therefore, mode is 15.

Now, we have to add four terms to this data such that mean and median remains same, but mode increase by 1.

So, mode of new data = 16

But as 16 is mode, its frequency must be greater than frequency of 2 i.e. minimum 3.

So, we have three terms out of four as 16, 16 and 16.

Now, let 4^th term be 'x'.

New data,

5, 8, 10, 15, 15, 16, 16, 16, 24, 28, x

No of terms = 11

As, mean remains same, mean of above data = 15

By using formula, mean of above data

⇒ 165 = 153 + x

⇒ x = 12

So, four terms are 12, 16, 16, 16.

Given,

No of observations, n = 10

Mean,

We know,

Arithmetic mean of x₁, x₂, x₃, …, x_n (n observations) is

For given data,

⇒ x₁ + x₂ + x₃ + …+x₁₀ = 200 [1]

Now, to find the mean of

x₁+4, x₂ + 8, …, x_n + 40

No of terms, n = 10

Mean using above formula,

Hence, mean of new data is 42.6

No of terms, n = 9

As, n is odd

Median is term.

⇒ Median is 5^th term

Let's first arrange given no in ascending order.

3, 5, 5, 7, 8, 9

Clearly, 9 is not possible as it is already at 6^th place and adding any term or terms can only shift its position ahead.

Also, 8 is possible if other three terms are greater than 8.

Therefore, 8 is the largest value possible.

No of terms, n = 9

As, n is odd

Median is term.

⇒ Median is 5^th term

If each of the largest 4 observations of set is increased by 2, it does not affect 5^th term or the order of terms.

Therefore, median remains same i.e. 20.

Next two class intervals 44 - 51 and 52 - 59.

(i) length of class interval = difference b/w upper and lower limit = 7

(ii)

(iii)

We know,

Class-mark

(i) As the difference between two consecutive class intervals is 12, we can evaluate the exclusive class intervals as,

Lower limit = class mark - 6

Upper limit = class mark + 6

(ii)

(iii)

For above problem, let us first calculate class marks for each interval.

Class mark

Now using points, we can make frequency curve and frequency polygon.

Frequency Polygon

Frequency Curve

Less than ogive

More than ogive