Statistics

The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies.

We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them.

The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies.

We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them.

Given the blood groups of 36 students can be represented in the form of frequency distribution table.

Just count the occurrence of blood groups in the data and formulate a table.

Count of blood group:

A = 10

B = 9

O = 15

AB = 2

Representing the same on a graph to show the relationship between blood group and number of students (frequency),

From the frequency table as well as from the graph, notice that:

Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15]

Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2]

Given the blood groups of 36 students can be represented in the form of frequency distribution table.

Just count the occurrence of blood groups in the data and formulate a table.

Count of blood group:

A = 10

B = 9

O = 15

AB = 2

Representing the same on a graph to show the relationship between blood group and number of students (frequency),

From the frequency table as well as from the graph, notice that:

Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15]

Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2]

Given the number of heads on tossing three coins can be represented in the form of frequency distribution table.

Just count the occurrence of number of heads in the data and formulate a table.

Count of occurrence of head by 3 coins:

0 (head occurred 0 times) = 3

1 (head occurred 1 times) = 10

2 (head occurred 2 times) = 10

3 (head occurred 3 times) = 7

It can also be analyzed and represented graphically:

Given the number of heads on tossing three coins can be represented in the form of frequency distribution table.

Just count the occurrence of number of heads in the data and formulate a table.

Count of occurrence of head by 3 coins:

0 (head occurred 0 times) = 3

1 (head occurred 1 times) = 10

2 (head occurred 2 times) = 10

3 (head occurred 3 times) = 7

It can also be analyzed and represented graphically:

Given the poll options (A, B & C) can be represented in the form of frequency distribution table.

Just count the occurrence of each option in the data and formulate a table.

Count of occurrence of options are:

A (complete prohibition) = 19

B (prohibition in public places only) = 36

C (not necessary) = 10

It can also be analyzed and represented graphically:

To find number of appropriate answers received in poll, add total frequency.

That is, 19 + 36 + 10 = 65

Thus, total number of appropriate answers received = 65

To analyze majority votes on poll, mark the highest frequency.

Note that, highest frequency is 36, that is of option B.

And hence, majority of people’s opinion is B, i.e., prohibition in public places only.

Given the poll options (A, B & C) can be represented in the form of frequency distribution table.

Just count the occurrence of each option in the data and formulate a table.

Count of occurrence of options are:

A (complete prohibition) = 19

B (prohibition in public places only) = 36

C (not necessary) = 10

It can also be analyzed and represented graphically:

To find number of appropriate answers received in poll, add total frequency.

That is, 19 + 36 + 10 = 65

Thus, total number of appropriate answers received = 65

To analyze majority votes on poll, mark the highest frequency.

Note that, highest frequency is 36, that is of option B.

And hence, majority of people’s opinion is B, i.e., prohibition in public places only.

Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.

There are 4 parameters for vehicles namely, cycles, autos, bikes and cars.

So now, create a table consisting of these vehicles and their corresponding frequencies.

Thus, this is the required frequency distribution table.

Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.

There are 4 parameters for vehicles namely, cycles, autos, bikes and cars.

So now, create a table consisting of these vehicles and their corresponding frequencies.

Thus, this is the required frequency distribution table.

Observe that, on x-axis we have classes from I Class to VI Class.

The scale used in x-axis is class on a unit interval.

And on y-axis, we have number of students.

The scale used on y-axis is from 0 to 90 with an interval of 10 units.

Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.

There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class.

So now, create a table consisting of these classes and their corresponding frequencies (number of students).

Thus, this is the required frequency distribution table.

Observe that, on x-axis we have classes from I Class to VI Class.

The scale used in x-axis is class on a unit interval.

And on y-axis, we have number of students.

The scale used on y-axis is from 0 to 90 with an interval of 10 units.

Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.

There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class.

So now, create a table consisting of these classes and their corresponding frequencies (number of students).

Thus, this is the required frequency distribution table.

Given are marks of 30 students of a class, obtained in a test (out of 75).

Note that these marks lie between 0 to 75.

So, these should be distributed in a way such that, the interval remains same in each distribution.

If we distribute it in a class interval equals to 5. Then, range would get maximized, i.e., 0-5, 5-10, 10-15, 15-20, …, 70-75.

But if we distribute it in a class interval of 10. Then the range will be optimized.

So, let’s make a table.

Thus, this is the required frequency distribution table.

Given are marks of 30 students of a class, obtained in a test (out of 75).

Note that these marks lie between 0 to 75.

So, these should be distributed in a way such that, the interval remains same in each distribution.

If we distribute it in a class interval equals to 5. Then, range would get maximized, i.e., 0-5, 5-10, 10-15, 15-20, …, 70-75.

But if we distribute it in a class interval of 10. Then the range will be optimized.

So, let’s make a table.

Thus, this is the required frequency distribution table.

Given are electricity bills (in rupees) of 25 houses in locality.

To construct a grouped frequency distribution table with a class size of 75, note down smallest and largest number from the data.

Smallest number = 170

Largest number = 724

So, let the class interval start from 150.

Then, add 75 to 150 = 150 + 75 = 225

⇒ First class interval = 150-225

Similarly,

Second class interval = 225-300 [∵ 225 + 75 = 300]

And so on…

For frequencies, count the occurrence of the numbers lying between the particular class interval.

Let us construct a table in order to show these class intervals along with their frequencies.

Thus, this is the required frequency distribution table.

Given are electricity bills (in rupees) of 25 houses in locality.

To construct a grouped frequency distribution table with a class size of 75, note down smallest and largest number from the data.

Smallest number = 170

Largest number = 724

So, let the class interval start from 150.

Then, add 75 to 150 = 150 + 75 = 225

⇒ First class interval = 150-225

Similarly,

Second class interval = 225-300 [∵, 225 + 75 = 300]

And so on…

For frequencies, count the occurrence of the numbers lying between the particular class interval.

Let us construct a table in order to show these class intervals along with their frequencies.

Thus, this is the required frequency distribution table.

Given are life (in years) of 40 batteries.

Know that, when the lower limit is included, but the upper limit is excluded, then it is an exclusive class interval. For example – 150-153, 153-156, etc … are exclusive type of class intervals. In the class interval 150 - 153, 150 is included but 153 is excluded.
Usually in the case of continuous variate, exclusive type of class intervals are used.

We can arrange the data in ascending order since the data is very large. We have

2.2, 2.3, 2.5, 2.6, 2.6, 2.8, 2.9, 2.9, 3.0, 3.0, 3.1, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.3, 3.4, 3.4, 3.4, 3.5, 3.5, 3.5, 3.5, 3.6, 3.7, 3.7, 3.7, 3.8, 3.8, 3.9, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.6

This makes calculation of frequencies easier.

So, let the class interval start from 2.0.

Then, add 0.5 to 2.0 = 2.0 + 0.5 = 2.5

⇒ First class interval = 2.0-2.5

Similarly,

Second class interval = 2.5-3.0 [∵ 2.5 + 0.5 = 3.0]

And so on…

For frequencies, count the occurrence of the numbers lying between the particular class interval.

Let us construct a table in order to show these class intervals along with their frequencies.

Thus, this is the required frequency distribution table.

Given are life (in years) of 40 batteries.

Know that, when the lower limit is included, but the upper limit is excluded, then it is an exclusive class interval. For example – 150-153, 153-156, etc … are exclusive type of class intervals. In the class interval 150 - 153, 150 is included but 153 is excluded.
Usually in the case of continuous variate, exclusive type of class intervals are used.

We can arrange the data in ascending order since the data is very large. We have

2.2, 2.3, 2.5, 2.6, 2.6, 2.8, 2.9, 2.9, 3.0, 3.0, 3.1, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.3, 3.4, 3.4, 3.4, 3.5, 3.5, 3.5, 3.5, 3.6, 3.7, 3.7, 3.7, 3.8, 3.8, 3.9, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.6

This makes calculation of frequencies easier.

So, let the class interval start from 2.0.

Then, add 0.5 to 2.0 = 2.0 + 0.5 = 2.5

⇒ First class interval = 2.0-2.5

Similarly,

Second class interval = 2.5-3.0 [∵, 2.5 + 0.5 = 3.0]

And so on…

For frequencies, count the occurrence of the numbers lying between the particular class interval.

Let us construct a table in order to show these class intervals along with their frequencies.

Thus, this is the required frequency distribution table.

Given is the frequency distribution table for ungrouped data.

We have

⇒ 25 number of parcels have weight each 50 kg,

34 number of parcels have weight each 65 kg,

And so on…

This also implies that,

Total weight of 25 parcels = 1250,

Total weight of 34 parcels = 2210,

And so on…

Mean of such ungrouped data is given by

⇒

⇒ [∵ ∑x_if_i = 17000 & ∑f_i = 200]

⇒ Mean = 85

Thus, mean weight of the parcel is 85 kg.

Given is the frequency distribution table for ungrouped data.

We have

⇒ 11 families each have no children,

25 families each have 1 child,

And so on…

This also implies that,

Total number of children of 11 families = 0 × 11 = 0,

Total number of children of 25 families = 1 × 25 = 25,

And so on…

Mean of such ungrouped data is given by

⇒

⇒ [∵ ∑x_if_i = 144 & ∑f_i = 84]

⇒ Mean = 1.7143

Thus, mean number of children per family is 1.7143.

Given is the frequency distribution table for ungrouped data.

We have

Mean of such ungrouped data is given by

⇒

⇒ [∵ Mean = 7.2]

⇒ 7.2(40 + K) = 260 + 10K

⇒ 288 + 7.2K = 260 + 10K

⇒ 10K – 7.2K = 288 – 260

⇒ 2.8K = 28

⇒ K = 28/2.8

⇒ K = 10

Thus, K = 10.

Given is the frequency distribution table for ungrouped data.

We have

⇒ 20 villages each have 12k population,

15 villages each have 5k population,

And so on…

This also implies that,

Total population of 20 villages = 12 × 20 = 240,

Total population of 15 villages = 5 × 15 = 75,

And so on…

Mean of such ungrouped data is given by

⇒

⇒ [∵ ∑x_if_i = 2571 & ∑f_i = 145]

⇒ Mean = 17.731

⇒ Mean = 17.731 × 1000

⇒ Mean = 17731

Thus, average population in each village is 17731.

Given is the frequency distribution table for ungrouped data.

We have

To find arithmetic mean of school wise savings in each mandal,

A common formula is:

In ‘Amberpet’:

⇒ Mean = Rs 359

In ‘Thriumalgiri’:

⇒ Mean = Rs 413

In ‘Saidabad’:

⇒ Mean = Rs 195

In ‘Khairathabad’:

⇒ Mean = Rs 228

In ‘Secundrabad’:

⇒ Mean = Rs 200

In ‘Bahadurpura’:

⇒ Mean = Rs 837

For arithmetic mean of savings of all the school:

Mean of such ungrouped data is given by

⇒

⇒ Mean = 444

Thus, arithmetic mean of savings of all the schools is Rs 444.

We have

We shall find median of the heights of the boys and girls separately.

Median of heights of boys:

N_xi = 37

⇒

Cumulative frequency = 19 falls in the height of 147 cm.

Thus, median height of boys is 147 cm.

Now, median of heights of girls:

N_yi = 29

⇒

Cumulative frequency = 15 falls in the height of 18.

Cumulative frequency, 18 falls in the height of 152 cm.

∴ Median height of boys is 147 cm whereas median height of girls in the class is 152 cm.

We have

For mean:

Mean is given by

⇒

⇒ Mean = 11.19

For median:

Total observation, N = 139

Since, N is odd (that is, 139 is odd),

⇒

⇒

Cumulative frequency = 79 is just greater than 70.

And cumulative frequency, 79 falls in no. of century, 10.

⇒ Median = 10

For mode:

Mode is the value having highest frequency in the data.

From the table given above, 56 is the greatest frequency in the given frequencies.

Thus, the corresponding value in ‘number of centuries’, 5 is the mode.

⇒ Mode = 5

We have

For mean:

Mean is given by

⇒

⇒ Mean = 80

For median:

Total observation, N = 150

Since, N is even (that is, 150 is even),

⇒

&

⇒

Cumulative frequencies = 88 is just greater than 75 and 76.

And cumulative frequencies, 75 and 76 falls in cost of packet, ‘75.

⇒ Median = 75

For mode:

Mode is the value having highest frequency in the data.

From the table given above, 36 is the greatest frequency in the given frequencies.

Thus, the corresponding value in ‘cost of packet’, 50 is the mode.

⇒ Mode = 50

Given: Mean of three students = 40 kg

The three students are Ranga, Rahim and Reshma, where

Ranga’s weight = 46 kg

Let Rahim’s weight = x

And Reshma’s weight = y [∵ Rahim’s weight = Reshma’s weight, according to the question]

Mean is given by

⇒ [∵ Mean = 40]

⇒ 120 = 46 + 2x

⇒ 2x = 120 – 46 = 74

⇒ x = 74/2

⇒ x = 37

Thus, Rahim’s weight = 37 kg

We have

For mean:

Mean is given by

⇒

⇒ Mean = 11.25

For median:

Total observation, N = 80

Since, N is even (that is, 80 is even),

⇒

&

⇒

Cumulative frequencies = 50 is just greater than 40 and 41.

And cumulative frequencies, 50 falls in donation by each student, 10.

⇒ Median = 10

For mode:

Mode is the value having highest frequency in the data.

From the table given above, 20 is the greatest frequency in the given frequencies.

Thus, the corresponding value in ‘donation by each student’, 10 is the mode.

⇒ Mode = 10

Given: There are 4 unknown numbers, but one of the number is 2.

Let x₁ = 2, x₂ = ?, x₃ = ? and x₄ = ?

According to the question, mean of first two numbers = 4

Mean of first two numbers is given by,

⇒

⇒

⇒ 8 = 2 + x₂

⇒ x₂ = 8 – 2 = 6

We have x₁ = 2 and x₂ = 6. …(i)

Also, according to the question, mean of first three numbers = 9

⇒

⇒ [from (i)]

⇒ 27 = 8 + x₃

⇒ x₃ = 27 – 8

⇒ x₃ = 19

We have now, x₁ = 2, x₂ = 6 and x₃ = 19 …(ii)

Also, according to the question, mean of all four numbers = 15

⇒

⇒ 60 = 2 + 6 + 19 + x₄ [from (ii)]

⇒ 60 = 27 + x₄

⇒x₄ = 60 – 27

⇒x₄ = 33

Thus, we have x₁ = 2, x₂ = 6, x₃ = 19 & x₄ = 33.