Statistics — Exercise 19.1

Find the mean of 8, 6, 10, 12, 1, 3, 4, 4.

Answer

Mean (M) = $\dfrac{\text{Sum of the all observations}}{\text{No. of observation}}$

Sum of all observations = 8 + 6 + 10 + 12 + 1 + 3 + 4 + 4 = 48.

$M = \dfrac{48}{8} = 6.$

Hence, mean = 6.

5 people were asked about the time in a week they spend in doing social work in their community. They replied 10, 7, 13, 20 and 15 hours, respectively. Find the mean time in a week devoted by them for social work.

Answer

Mean (M) = $\dfrac{\text{Total social work hrs}}{\text{No. of people}}$

Sum of hours spent by people on social work = 10 + 7 + 13 + 20 + 15 = 65.

$M = \dfrac{65}{5} = 13.$

Hence, the mean time in a week devoted by people for social work is 13 hours.

The enrollment of a school during six consecutive years was as follows:

1620, 2060, 2540, 3250, 3500, 3710.

Find the mean enrollment.

Answer

Mean (M) = $\dfrac{\text{Sum of enrollments}}{\text{No. of years}}$

Sum of enrollments = 1620 + 2060 + 2540 + 3250 + 3500 + 3710 = 16680

$M = \dfrac{16680}{6} = 2780.$

Hence, the mean enrollment = 2780.

Find the mean of the first twelve natural numbers.

Answer

Mean (M) = $\dfrac{\text{Sum of first 12 natural nos.}}{\text{No. of natural nos.}}$

Sum of first 12 natural numbers = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78.

$M = \dfrac{78}{12} = 6.5$

Hence, the mean of the first twelve natural numbers = 6.5

Find the mean of the first six prime numbers.

Answer

First six prime numbers = 2, 3, 5, 7, 11, 13.

Mean (M) = $\dfrac{\text{Sum of prime numbers}}{\text{No. of prime numbers}}$

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

$M = \dfrac{41}{6}$.

Hence, the mean of the first six prime numbers = $\dfrac{41}{6}$.

Find the mean of the first seven odd prime numbers.

Answer

First seven odd prime numbers = 3, 5, 7, 11, 13, 17, 19.

Mean (M) = $\dfrac{\text{Sum of first seven odd prime nos.}}{\text{No. of prime nos.}}$

Sum of first seven odd prime numbers = 3 + 5 + 7 + 11 + 13 + 17 + 19 = 75.

$M = \dfrac{75}{7} \\[1em] = 10\dfrac{5}{7}.$

Hence, the mean of the first seven odd prime numbers = $10\dfrac{5}{7}$.

The marks (out of 100) obtained by a group of students in a Mathematics test are 81, 72, 90, 90, 85, 86, 70, 93 and 71. Find the mean marks obtained by the group of students.

Answer

Mean marks (M) = $\dfrac{\text{Total marks}}{\text{No. of students}}$

Total marks = 81 + 72 + 90 + 90 + 85 + 86 + 70 + 93 + 71 = 738.

$M = \dfrac{738}{9} = 82.$

Hence, mean marks = 82.

The mean of the age of three students Vijay, Rahul and Rakhi is 15 years. If their ages are in the ratio 4 : 5 : 6 respectively, then find their ages.

Answer

Let age of Vijay, Rahul and Rakhi be 4x, 5x and 6x respectively..

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

Sum of age of students = 4x + 5x + 6x = 15x.

$\therefore 15 = \dfrac{15x}{3} \\[1em] \Rightarrow 15 \times 3 = 15x \\[1em] \Rightarrow x = \dfrac{15 \times 3}{15} \\[1em] \Rightarrow x = 3.$

Vijay's age = 4x = 4 × 3 = 12 years,

Rahul's age = 5x = 5 × 3 = 15 years,

Rakhi's age = 6x = 6 × 3 = 18 years.

Hence, age of three students Vijay, Rahul and Rakhi are 12, 15 and 18 years respectively.

The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.

Answer

Given, mean of 5 numbers = 20.

By formula,

Mean = $\dfrac{\text{Total sum of observations}}{\text{No. of observations}}$

$\therefore 20 = \dfrac{\text{Total sum of observations}}{5} \\[1em] \text{Total sum of observations} = 20 \times 5 = 100.$

Let number excluded be x.

Given, Mean of the remaining numbers becomes 23.

$\therefore 23 = \dfrac{100 - x}{4} \\[1em] \Rightarrow 23 \times 4 = 100 - x \\[1em] \Rightarrow 92 = 100 - x \\[1em] \Rightarrow x = 100 - 92 \\[1em] \Rightarrow x = 8.$

Hence, excluded number = 8.

The mean of 25 observations is 27. If one observation is included, the mean still remains 27. Find the included observation.

Answer

Given,

The mean of 25 observations is 27.

By formula,

Mean = $\dfrac{\text{Total sum of observations}}{\text{No. of observations}}$

$\therefore 27 = \dfrac{\text{Total sum of observations}}{25} \\[1em] \text{Total sum of observations} = 27 \times 25 = 675.$

Given, mean after including new number remains 27.

No. of observations = 25 + 1 = 26.

$\therefore 27 = \dfrac{\text{New sum of observations}}{26} \\[1em] \text{New sum of observations} = 27 \times 26 = 702.$

New number included = New sum of observations - Total sum of observations = 702 - 675 = 27.

Hence, the new included observation = 27.

The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.

Answer

By formula,

Mean = $\dfrac{\text{Total sum of observations}}{\text{No. of observations}}$

Given,

The mean of 5 observations = 15

$\therefore 15 = \dfrac{\text{Total sum of 5 observations}}{5} \\[1em] \Rightarrow \text{Total sum of observations} = 15 \times 5 \\[1em] \Rightarrow \text{Total sum of observations} = 75.$

Given,

Mean of first 3 observations = 14

$\therefore 14 = \dfrac{\text{Sum of first 3 observations}}{3} \\[1em] \Rightarrow \text{Sum of first three observations} = 14 \times 3 \\[1em] \Rightarrow \text{Sum of first three observations} = 42.$

Given,

Mean of last 3 observations = 17

$\therefore 17 = \dfrac{\text{Sum of last 3 observations}}{3} \\[1em] \Rightarrow \text{Sum of last three observations} = 17 \times 3 \\[1em] \Rightarrow \text{Sum of last three observations} = 51.$

Third observation = Sum of first three observations + Sum of last three observations - Total sum of observation
= 51 + 42 - 75 = 18.

Hence, the third observation = 18.

The mean of 8 variates is 10.5. If seven of them are 3, 15, 7, 19, 2, 17 and 8, then find the 8th variate.

Answer

By formula,

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

Given,

Mean of 8 variates = 10.5

$\therefore 10.5 = \dfrac{\text{Sum of eight variates}}{8} \\[1em] \text{Sum of 8 variates} = 10.5 \times 8 = 84.$

Given,

Seven variates = 3, 15, 7, 19, 2, 17 and 8.

Sum of seven variates = (3 + 15 + 7 + 19 + 2 + 17 + 8) = 71.

8th variate = Sum of 8 variates - Sum of 7 variates = 84 - 71 = 13.

Hence, 8th variate = 13.

The mean weight of 8 students is 45.5 kg. Two more students having weights 41.7 kg and 53.3 kg join the group. What is the new mean weight?

Answer

By formula,

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

Given,

$\therefore 45.5 = \dfrac{\text{Total weight of eight students}}{8} \\[1em] \text{Total weight of eight students} = 45.5 \times 8 = 364.$

Weight of two more students are 41.7 kg and 53.3 kg

Now,

The total weight of 10 students = 364 + 41.7 + 53.3 = 459 kg.

New mean weight = $\dfrac{459}{10}$ = 45.9 kg.

Hence, new mean weight = 45.9 kg.

Mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.

Answer

By formula,

Mean = $\dfrac{\text{Total sum of observations}}{\text{No. of observations}}$

$\therefore 35 = \dfrac{\text{Total sum of observations}}{9} \\[1em] \text{Total sum of observations} = 35 \times 9 = 315.$

Since, 81 was misread as 18.

So, Actual sum of observation = Total sum of observations - 18 + 81 = 315 - 18 + 81 = 378.

Correct mean = $\dfrac{378}{9}$ = 42.

Hence, the correct mean = 42.

A student scored the following marks in 11 questions of a question paper:

7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6.

Find the median marks.

Answer

Given,

Marks scored in 11 questions of a question paper by the student are:

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

Arranging it in ascending order, we have

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

Here, n = 11 which is odd.

∴ Median = $\dfrac{(n + 1)}{2}$ th term

= $\dfrac{11 + 1}{2} = \dfrac{12}{2}$ = 6th term.

6th term = 5.

Hence, median marks = 5.

Calculate the mean and the median of the numbers:

2, 3, 4, 3, 0, 5, 1, 1, 3, 2.

Answer

By formula,

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

Sum of numbers = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24

$\therefore \text{Mean } = \dfrac{24}{10} \\[1em] = 2.4$

Given,

Numbers : 2, 3, 4, 3, 0, 5, 1, 1, 3, 2.

First arrange the numbers in ascending order :

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

Here, n = 10 which is even.

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{ th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2}\text{ th observation} + \Big(\dfrac{10}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \dfrac{5 \text{ th observation} + 6\text{ th observation}}{2} \\[1em] = \dfrac{2 + 3}{2} \\[1em] = \dfrac{5}{2} \\[1em] = 2.5$

Hence, mean = 2.4 and median = 2.5

A group of students was given a special test in Mathematics. The test was completed by the various students in the following time in (minutes) :

24, 30, 28, 17, 22, 36, 30, 19, 32, 18, 20, 24.

Find the mean time and median time taken by the students to complete the test.

Answer

By formula,

$\text{Mean} = \dfrac{\text{Total minutes}}{\text{No. of students}} \\[1em] = \dfrac{24 + 30 + 28 + 17 + 22 + 36 + 30 + 19 + 32 + 18 + 20 + 24}{12} \\[1em] = \dfrac{300}{12} \\[1em] = 25.$

Arranging the data in ascending order :

17, 18, 19, 20, 22, 24, 24, 28, 30, 30, 32, 36

Here, n = 12 which is even

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{ th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{12}{2}\text{ th observation} + \Big(\dfrac{12}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \dfrac{6 \text{ th observation} + 7\text{ th observation}}{2} \\[1em] = \dfrac{24 + 24}{2} \\[1em] = \dfrac{48}{2} \\[1em] = 24$

Hence, mean time = 25 minutes and median time = 24 minutes.

In a Science test given to a group of students, the marks scored by them (out of 100) are

41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.

Find the mean and median of this data.

Answer

By formula,

$\text{Mean} = \dfrac{\text{Total marks}}{\text{No. of students}}$

Total marks = 41 + 39 + 52 + 48 + 54 + 62 + 46 + 52 + 40 + 96 + 42 + 40 + 98 + 60 + 52
= 822

$\text{Mean} = \dfrac{822}{15} \\[1em] = 54.8$

Given,

Student's marks : 41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.

On arranging the marks obtained by the students in ascending order, we have :

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

Here, n = 15 which is odd.

By formula,

$\text{Median } = \dfrac{n + 1}{2} \text{ th observation} \\[1em] = \dfrac{15 + 1}{2} \text{ th observation} \\[1em] = \dfrac{16}{2} \text{ th observation} \\[1em] = 8 \text{ th observation} \\[1em] = 52.$

Hence, mean = 54.8 and median = 52.

The points scored by a Kabaddi team in a series of matches are as follows:

7, 17, 2, 5, 27, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18.

Find the mean and the median of the points scored by the Kabaddi team.

Answer

Given,

Points scored by kabaddi team : 7, 17, 2, 5, 27, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18.

By formula,

$\text{Mean} = \dfrac{\text{Sum of points}}{\text{No. of matches}} \\[1em] = \dfrac{7 + 17 + 2 + 5 + 27+ 15 + 8 + 14 + 10 + 48 + 10 + 7 + 24 + 8 + 28 + 18}{16} \\[1em] = \dfrac{248}{16} \\[1em] = 15.5$

Let’s arrange the given data in ascending order:

2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48.

Here, n = 16 when is even

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{ th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{16}{2}\text{ th observation} + \Big(\dfrac{16}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \dfrac{8 \text{ th observation} + 9\text{ th observation}}{2} \\[1em] = \dfrac{10 + 14}{2} \\[1em] = \dfrac{24}{2} \\[1em] = 12$

Hence, the mean and the median of the points scored by the Kabaddi team are 15.5 and 12 respectively.

The following observations have been arranged in ascending order. If the median of the data is 47.5, find the value of x.

17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93.

Answer

Given data,

17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93.

Here, n = 14 which is even.

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{ th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{14}{2}\text{ th observation} + \Big(\dfrac{14}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \dfrac{7 \text{ th observation} + 8\text{ th observation}}{2} \\[1em] = \dfrac{x + 50}{2}.$

Given, median = 47.5

⇒ $\dfrac{x + 50}{2}$ = 47.5

⇒ x + 50 = 95

⇒ x = 95 - 50

⇒ x = 45.

Hence, the value of x is 45.

The following observations have been arranged in ascending order. If the median of the data is 13, find the value of x.

3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.

Answer

Given observations in ascending order :

3, 6, 7, 10, x, x + 4, 19, 20, 25, 28

Here, n = 10 which is even.

By formula,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{ th observation} + \Big(\dfrac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2}\text{ th observation} + \Big(\dfrac{10}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \dfrac{5 \text{ th observation} + 6\text{ th observation}}{2} \\[1em] = \dfrac{x + x + 4}{2} \\[1em] = \dfrac{2x + 4}{2} \\[1em] = x + 2.$

Given, median = 13

⇒ x + 2 = 13

⇒ x = 13 - 2

⇒ x = 11.

Hence, the value of x is 11.