Statistics

Note that, class intervals have varying base widths. This changes how we generally draw a histogram. And there is a need to find length of rectangles for each class interval because length of rectangle will determine y-axis now, while class interval will be represented on the x-axis.

Length of the rectangle is given by,

Where C = minimum class width of the data set.

Here, note from the table below that C = 5.

Thus, histogram can be drawn following the given table. Table makes our observation and calculation clearer and ensures accuracy of the final answer.

The histogram is as follows:

Note that, class intervals have varying base widths. This changes how we generally draw a histogram.

There is a need to find length of rectangles for each class interval because length of rectangle will determine y-axis now, while class interval will be represented on the x-axis.

Length of the rectangle is given by,

Where C = minimum class width of the data set.

Here, note from the table below that C = 200.

Thus, histogram can be drawn following the given table. Table makes our observation and calculation clearer and ensures accuracy of the final answer.

The histogram is as follows:

Note that, class intervals have varying base widths as well as inclusive type of data. This changes how we generally draw a histogram.

We need to convert the data into exclusive type and find length of rectangles for each class interval because length of rectangle will determine y-axis now, while class interval will be represented on the x-axis.

Length of the rectangle is given by,

Where C = minimum class width of the data set.

Here, note from the table below that C = 4.

Thus, histogram can be drawn following the given table. Table makes our observation and calculation clearer and ensures accuracy of the final answer.

To convert the inclusive type of data into exclusive type of data, subtract 0.5 from each lower value of the class interval and add 0.5 to each upper value of the class interval.

The histogram is as follows:

Note that, class intervals have varying base widths. This changes how we generally draw a histogram.

There is a need to find length of rectangles for each class interval because length of rectangle will determine y-axis now, while class interval will be represented on the x-axis.

Length of the rectangle is given by,

Where C = minimum class width of the data set.

Here, note from the table below that C = 4.

Thus, histogram can be drawn following the given table. Table makes our observation and calculation clearer and ensures accuracy of the final answer.

The histogram is as follows:

Note that, class intervals have varying base widths. This changes how we generally draw a histogram.

There is a need to find length of rectangles for each class interval because length of rectangle will determine y-axis now, while class interval will be represented on the x-axis.

Length of the rectangle is given by,

Where C = minimum class width of the data set.

Here, note from the table below that C = 10.

Thus, histogram can be drawn following the given table. Table makes our observation and calculation clearer and ensures accuracy of the final answer.

The histogram is as follows:

The data given here is raw data of the number of bags sold in their respective days.

Mean of such data is given by

Where, ∑x = Sum of all observations

& n = Total number of observations

So,

⇒

Thus, mean number of bags sold by the shopkeeper on the given days is 28.67.

Given is the data of children present in 10 families in a locality.

Mean of number of children in a family = 4

Mean is given by

Where ∑x = Sum of all observations

& n = Total number of observations.

Here, n = 10

So,

⇒

⇒ 34 + x = 4 × 10

⇒ 34 + x = 40

⇒ x = 40 – 34

⇒ x = 6

Thus, x = 6.

Given that, mean of 20 numbers = 59

Mean can be given by

Where, ∑x = Sum of all observations

& n = Total number of observations

Here, Mean = 59 and n = 20.

So,

⇒

⇒ ∑x = 59 × 20

⇒ ∑x = 1180

We have sum of all observations = 1180

If 3 is added to each of the 20 numbers, then new sum of all observations = 1180 + (20 × 3)

⇒ New sum of all observations = 1180 + 60 = 1240

Now, we have fresh data: ∑x’ = 1240, n = 20

⇒ New mean = 62

Thus, new mean of the observation is 62.

Given that, mean of 15 numbers = 44

Mean can be given by

Where, ∑x = Sum of all observations

& n = Total number of observations

Here, Mean = 44 and n = 15.

So,

⇒

⇒ ∑x = 44 × 15

⇒ ∑x = 660

We have sum of all observations = 660

If 7 is subtracted from each of the 15 numbers, then the new sum of all observations = 660 – (15 × 7)

⇒ New sum of all observations = 660 – 105 = 555

Now, we have fresh data: ∑x’ = 555, n = 15

⇒ New mean = 37

Thus, new mean of the observation is 37.

Given that, mean of 12 numbers = 48

Mean can be given by

Where, ∑x = Sum of all observations

& n = Total number of observations

Here, Mean = 48 and n = 12.

So,

⇒

⇒ ∑x = 48 × 12

⇒ ∑x = 576

We have sum of all observations = 576

If 4 is multiplied by each of the 12 numbers, then the new sum of all observations = 576 × 4

⇒ New sum of all observations = 2304

Now, we have fresh data: ∑x’ = 2304, n = 12

⇒ New mean = 192

Thus, new mean of the observation is 192.

Given that, mean of 16 numbers = 54

Mean can be given by

Where, ∑x = Sum of all observations

& n = Total number of observations

Here, Mean = 54 and n = 16.

So,

⇒

⇒ ∑x = 54 × 16

⇒ ∑x = 864

We have sum of all observations = 864

If 9 is divided from each of the 16 numbers, then the new sum of all observations = 864 ÷ 9

⇒ New sum of all observations = 96

Now, we have fresh data: ∑x’ = 96, n = 16

⇒ New mean = 6

Thus, new mean of the observation is 6.

Given is that, mean weight of 6 boys = 48 kg

Mean is given by,

Where, ∑x = sum of all observations

& n = total number of observations

Here, n = 6 and mean = 48.

Let the weight of 6^th boy be x.

So,

⇒

⇒ 227 + x = 48 × 6

⇒ 227 + x = 288

⇒ x = 288 – 227

⇒ x = 61

Thus, weight of the sixth boy is 61 kg.

In assumed-mean method, we need to assume a mean value (the near to middle or middlemost value) from x_i’s.

We have the following table.

In assumed mean method, arithmetic mean is given by

Where, A = assumed mean = 53

∑f_id_i = -17

& ∑f_i = 40

Putting all these value in the formula, we get

⇒

⇒

Thus, mean weight of 40 students is 52.575 kg.

Given that, the total number of observations = 75

Mean of 75 observations = 27

Mean is given by,

⇒ Sum of all observation = Mean × Total number of observations

⇒ Sum of all observation = 27 × 75 = 2025

Now, according to the question, one of the value in the observation was wrongly taken as 43 instead of 53.

So, in order to obtain the new mean, we need to subtract 43 from the sum of observations and add the correct value, 53 to it.

⇒

⇒

Thus, the new mean = 27.13

Given that, the total number of observations = 100

Mean of 100 observations = 40

Mean is given by,

⇒ Sum of all observation = Mean × Total number of observations

⇒ Sum of all observation = 40 × 100 = 4000

Now, according to the question, two items in the observation were wrongly taken as 30 and 27 instead of 3 and 72.

So, in order to obtain the new mean, we need to subtract 30 and 27 from the sum of observations and add the correct values, 3 and 72 to it.

⇒

⇒

Thus, the new mean = 40.18

We can make the following table for ease of calculation:

Mean is given by

⇒

⇒ Mean = 28.67

Thus, the average number of patients attending the hospital are 28.67.

In step deviation method, we divide deviation by the width of the class intervals in order to simplify the calculation.

Here, we shall assume a mean from the given x_i’s. Let the assumed mean (A) be 35.

Also,

Where, A = assumed mean = (here) 35

& c = width of the class intervals = (here) 10

Construct the following table:

Using these values, we can find the arithmetic mean of the following data.

Arithmetic mean is given by

⇒

⇒ Mean = 35 – 7

⇒ Mean = 28

Thus, arithmetic mean of the following data is 28.

Let’s convert the inclusive type of data into exclusive type. We can do this by subtracting 0.5 from each lower limits of the class intervals and by adding 0.5 to each upper limits of the class interval.

Using direct method, we can find arithmetic mean of the data.

Let’s make the following table for ease of calculation:

Mean is given by

⇒

⇒ Mean = 48.1

Thus, the arithmetic mean of the data is 48.1.

In step deviation method, we divide deviation by the width of the class intervals in order to simplify the calculation.

Here, we shall assume a mean from the given x_i’s. Let the assumed mean (A) be 325.

Also,

Where, A = assumed mean = (here) 325

& c = width of the class intervals = (here) 50

Construct the following table:

Using these values, we can find the arithmetic mean of the following data.

Arithmetic mean is given by

⇒

⇒

⇒ Mean = 326.25

Thus, arithmetic mean of the following data is 326.25.

Let’s convert the inclusive type of data into exclusive type. We can do this by subtracting 0.5 from each lower limits of the class intervals and by adding 0.5 to each upper limits of the class interval.

Using direct method, we can find arithmetic mean of the data.

Let’s make the following table for ease of calculation:

Mean is given by

⇒

⇒ Mean = 55.5

Thus, the arithmetic mean of the data is 55.5.

(i). Let us arrange the numbers in ascending order. (Without arranging the values in ascending/descending order, we will not be able to get the correct median) We have,

12, 18, 32, 51, 58, 92 ,106 …(1)

Total number of items, n = 7 (odd number).

When n is odd, median is given by

⇒

⇒

⇒ Median = 4^th term

Count 4^th term from the arranged numbers in (1).

The 4^th term = 51

⇒ Median = 51 [∵ Median = 4^th term]

Thus, median is 51.

(ii). Let us arrange the numbers in ascending order. (Without arranging the values in ascending/descending order, we will not be able to get the correct median) We have,

1, 2, 3, 7, 9, 14, 15, 18, 21, 28, 46, 59 …(1)

Total number of items, n = 12 (even number).

When n is even, median is given by

⇒

⇒ Median = mean of 6^th and 7^th term

Count 6^th and 7^th term from the arranged numbers in (1).

The 6^th term = 14

& 7^th term = 15

⇒ Median = mean of 14 and 15 …(2)

Mean of 14 and 15 is calculated as,

⇒

Put mean of 14 and 15, 14.5 in equation (2),

Median = 14.5

Thus, median is 14.5.

Since, we have ungrouped data, we need to sort the data in ascending order to find median of the data.

Let us formulate a table for easy and simple calculations.

Total number of items, n = 20 (even)

When n = even, position of median is given by

⇒

⇒

Note that, from the cumulative frequency column: 10^th value = 15 & 11^th value = 19

Then,

⇒

⇒ Median = 17

Thus, median is 17.

This is a grouped frequency distribution, in which median is calculated as follows.

We have the following table:

Here, we get total frequency, N = 30. So,

Since, N/2 = 15. Mark a cumulative frequency which is just greater than 15.

Here, it is 17. So, median class would be 15 – 20.

Median is given by

Where, l = lower limit of the median class = 15

N/2 = 15 (as calculated above)

m = cumulative frequency of the class preceding the median class = 7

f = frequency of the median class = 10

c = width of the median class = 5

Putting all these values in the formula, we get

⇒

⇒ Median = 15 + 4 = 19

Thus, median is 19.

This is a grouped frequency distribution, in which median is calculated as follows.

First, we need to convert this inclusive type of data values into exclusive type. For this, subtract 0.5 from lower limits of each class and add 0.5 to upper limits of each class interval.

We have the following table:

Here, we get total frequency, N = 50. So,

Since, N/2 = 25. Mark a cumulative frequency which is just greater than 25.

Here, it is 31. So, median class would be 29.5 – 39.5.

Median is given by

Where, l = lower limit of the median class = 29.5

N/2 = 25 (as calculated above)

m = cumulative frequency of the class preceding the median class = 20

f = frequency of the median class = 11

c = width of the median class = 10

Putting all these values in the formula, we get

⇒

⇒ Median = 29.5 + 4.55 = 34.05

Thus, median is 34.05.

This is a grouped frequency distribution, in which median is calculated as follows.

First, we need to convert this inclusive type of data values into exclusive type. For this, subtract 0.5 from lower limits of each class and add 0.5 to upper limits of each class interval.

We have the following table:

Here, we get total frequency, N = 80. So,

Since, N/2 = 40. Mark a cumulative frequency which is just greater than 40.

Here, it is 44. So, median class would be 10.5 – 15.5.

Median is given by

Where, l = lower limit of the median class = 10.5

N/2 = 40 (as calculated above)

m = cumulative frequency of the class preceding the median class = 19

f = frequency of the median class = 25

c = width of the median class = 5

Putting all these values in the formula, we get

⇒

⇒ Median = 10.5 + 4.2 = 14.7

Thus, median is 14.7.

This is a grouped frequency distribution, in which median is calculated as follows.

We have the following table:

Here, we get total frequency, N = 800. So,

Since, N/2 = 400. Mark a cumulative frequency which is just greater than 400.

Here, it is 550. So, median class would be 40 – 45.

Median is given by

Where, l = lower limit of the median class = 40

N/2 = 400 (as calculated above)

m = cumulative frequency of the class preceding the median class = 400

f = frequency of the median class = 150

c = width of the median class = 5

Putting all these values in the formula, we get

⇒

⇒ Median = 40 + 0 = 40

Thus, median is 40.

Mode is the number which has highest frequency, that is, largest occurrence.

For ease of observation, we can arrange the data in ascending order. We have,

42, 45, 47, 47, 49, 52, 65, 65, 71, 71, 72, 72, 72, 75, 82

This can be arranged in a tabular form showing frequencies of marks:

The highest frequency is 3, and its corresponding mark is 72.

Thus, modal mark is 72.

Mode is nothing but a number with highest frequency. Interpret the column ‘No. of pairs sold’ in the terms of frequency.

The highest frequency is 21, and its corresponding size of shoe is 7.

Thus, mode of the following data is 7.

This is a grouped frequency distribution. To find mode of this type of data, we can interpret the column ‘No. of patients’ as frequency.

Note that, highest frequency = 50.

Corresponding class interval = 40 – 50

⇒ Modal class = 40 – 50

Mode is given by,

Where, l = lower limit of the modal class = 40

f = frequency of the modal class = 50

f₁ = frequency of the class preceding the modal class = 36

f₂ = frequency of the class succeeding the modal class = 20

c = width of the class interval = 10

Putting all these values in the modal formula,

⇒

⇒

⇒

⇒ Mode = 43.18

Thus, the mode is 43.18.

This is a grouped frequency distribution and an inclusive type of data.

To find mode of this type of data, we must convert this inclusive type of data into exclusive type. For exclusive type data, we need to subtract 0.5 from lower limits of each class interval and add 0.5 to upper limits of each class intervals.

Also, we can interpret the column ‘No. of students’ as frequency.

Note that, highest frequency = 20

Corresponding class interval = 40.5 – 45.5

⇒ Modal class = 40.5 – 45.5

Mode is given by,

Where, l = lower limit of the modal class = 40.5

f = frequency of the modal class = 20

f₁ = frequency of the class preceding the modal class = 18

f₂ = frequency of the class succeeding the modal class = 14

c = width of the class interval = 5

Putting all these values in the modal formula,

⇒

⇒

⇒ Mode = 40.5 + 1.25

⇒ Mode = 41.75

Thus, the mode is 41.75.

For Mean:

Since, this is raw data. Mean is given by,

⇒

⇒

⇒ Mean = 14

Thus, mean of the data is 14.

For Median:

We need to arrange the raw data in ascending order. We have,

13, 13, 13, 13, 14, 14, 15, 15, 15, 15

Total number of items, n = 10 (even)

When n is even, then median is given by

…(i)

Calculate (n/2)^th position.

⇒

⇒

[∵ from the arranged data, 5^th term = 14] …(ii)

Now, calculate (n/2 + 1)^th position.

⇒

⇒

[∵ from the arranged data, 6^th term = 14] …(iii)

Putting (ii) and (iii) in equation (i), we get

Median = Mean of 14 and 14

⇒

⇒

⇒ Median = 14

Thus, median is 14.

For Mode:

We need to arrange the raw data in ascending order for ease of observation. We have,

13, 13, 13, 13, 14, 14, 15, 15, 15, 15

From observing from the arranged data, note that

13 → 4 times

14 → 2 times

15 → 4 times

Representing it in tabular form,

Highest frequency = 4

And the corresponding data are 13 and 15.

This is a bimodal distribution, because it has two modes.

Thus, modes are 13 and 15.

We have

For Mean:

Mean is given by

⇒ [from the table]

⇒ Mean = 4.05

For Median:

Total number of items, n = 5 (odd number).

When n is odd, median is given by

⇒

⇒

⇒ Median = 3^rd term

Count 3^rd term from the arranged numbers in the table.

The 3^rd term = 4

⇒ Median = 4 [∵ Median = 3^rd term]

For Mode:

Mode is nothing but a number with highest frequency. Interpret the column ‘No. of plants’ in the terms of frequency.

Here, highest frequency is 28. Then, corresponding no. of branches are 4.

⇒ Mode = 4

Thus, mean is 4.05, median is 4 and mode is 4.

We have

For Mean:

Mean is given by

⇒ [from the table]

⇒ Mean = 32.1

For Median:

Here, we get total frequency, ∑f_i = N = 50. So,

Since, N/2 = 25. Mark a cumulative frequency which is just greater than 25.

Here, it is 29. So, median class would be 24.5 – 34.5.

Median is given by

Where, l = lower limit of the median class = 24.5

N/2 = 25 (as calculated above)

m = cumulative frequency of the class preceding the median class = 17

f = frequency of the median class = 12

c = width of the median class = 10

Putting all these values in the formula, we get

⇒

⇒ Median = 24.5 + 6.67 = 31.167

For Mode:

We have the following data in exclusive type in tabular form. We can interpret the column ‘No. of cases’ as frequency.

Note that, highest frequency = 12

Corresponding class interval = 24.5 – 34.5

⇒ Modal class = 24.5 – 34.5

Mode is given by,

Where, l = lower limit of the modal class = 24.5

f = frequency of the modal class = 12

f₁ = frequency of the class preceding the modal class = 11

f₂ = frequency of the class succeeding the modal class = 10

c = width of the class interval = 10

Putting all these values in the modal formula,

⇒

⇒

⇒ Mode = 24.5 + 3.33

⇒ Mode = 27.83

Thus, mean is 32.1, median is 32.167 and mode is 27.83.

We have

For Mean:

Mean is given by

⇒ [from the table]

⇒ Mean = 28

For Median:

Here, we get total frequency, ∑f_i = N = 20. So,

Since, N/2 = 10. Mark a cumulative frequency which is just greater than 10.

Here, it is 18. So, median class would be 30 – 40.

Median is given by

Where, l = lower limit of the median class = 30

N/2 = 10 (as calculated above)

m = cumulative frequency of the class preceding the median class = 10

f = frequency of the median class = 8

c = width of the median class = 10

Putting all these values in the formula, we get

⇒

⇒ Median = 30 + 0 = 30

For Mode:

We have the following data in exclusive type in tabular form. We can interpret the column ‘No. of students’ as frequency.

Note that, highest frequency = 8

Corresponding class interval = 30 – 40

⇒ Modal class = 30 – 40

Mode is given by,

Where, l = lower limit of the modal class = 30

f = frequency of the modal class = 8

f₁ = frequency of the class preceding the modal class = 5

f₂ = frequency of the class succeeding the modal class = 2

c = width of the class interval = 10

Putting all these values in the modal formula,

⇒

⇒

⇒ Mode = 30 + 3.33

⇒ Mode = 33.3

Thus, mean is 28, median is 30 and mode is 33.3.

Let us list out first 10 natural numbers. They are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Total number of observations = 10

Mean is given by

⇒

⇒

⇒ Mean = 5.5

Thus, option (C) is correct.

The integers from -5 and 5 are:

-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Total number of observations = 11

Mean is given by

⇒

⇒ Mean = 0

Thus, option (B) is correct.

Here, total number of observations = 5

And given that, mean = 20

Mean is given by

⇒

⇒ 5x + 20 = 5 × 20

⇒ 5x + 20 = 100

⇒ 5x = 100 – 20

⇒ 5x = 80

⇒

⇒ x = 16

Thus, option (B) is correct.

We can arrange the data in ascending form as:

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5

This gives us an ease of observation.

Here, 5 is coming 5 times and it is the most occurring value in the data.

Since, mode is a number having the highest frequency of all numbers in the data.

⇒ Mode = 5

Thus, option (D) is correct.

To find median, we need to arrange data in ascending order.

We have

9, 10, 11 ,12, 14

Total number of observations, n = 5 (odd).

When n is odd, median is found as follows.

⇒

⇒

⇒ Median = 3^rd term

⇒ Median = 11 [∵ the 3^rd term in the arranged data = 11]

Thus, option (A) is correct.

To find median, we need to arrange data in ascending order.

We have

1, 2, 4, 7, 8, 9

Total number of observations, n = 6 (even).

When n is even, median is found as follows.

⇒

⇒ Median = Mean of 3^rd term and 4^th term

⇒ Median = Mean of 4 and 7 [∵ from the arranged data, 3^rd term = 4 and 4^th term = 7]

⇒

⇒

⇒ Median = 5.5

Thus, option (C) is correct.

The first whole numbers are: 0, 1, 2, 3, 4

Total number of observations = 5

Mean is given by

⇒

⇒

⇒ Mean = 2

Thus, option (A) is correct.

Given that, arithmetic mean of 10 numbers = -7

We know, total number of observations, n = 10

Then, mean is given by

⇒

⇒ Sum of all observations = 10 × -7 = -70

According to question, if 5 is added to every 10 number then,

New sum of all observations = Old sum + (5 × 10)

⇒ New sum = -70 + 50 = -20

⇒

⇒ New mean = -2

Thus, option (A) is correct.

All the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. [These numbers divide 24 completely]

Total number of observations = 8

Mean is given by

⇒

⇒ Mean = 7.5

Thus, option (D) is correct.

Given that, Total number of observations = 5

Mean = 20

So, Sum of observations can be given as

Sum of observations = Mean × Total number of observations [∵ ]

⇒ Sum of observations = 20 × 5

⇒ Sum of observations = 100

Let the excluded number be x.

New Sum of observations = 100 – x

New Mean = 15

New total number of observations = 4

Mean is given by

⇒

⇒ 100 – x = 15 × 4

⇒ 100 – x = 60

⇒ x = 100 – 60 = 40

⇒ The excluded number is 40.

Thus, option (B) is correct.