1. Find the range of heights of any ten students in your class.
Solution:-
The heights (in cm) of \(10\) students in our class \(= 130, 132, 135, 137, 139, 140, 142, 143, 145, 148\)
By observing the above-mentioned values, the highest value is \(= 148\) cm
By observing the above-mentioned values, the lowest value is \(= 130\) cm
Range of Heights = Highest value – Lowest value \(= 148 – 130 = 18\) cm
Solution:-
The heights (in cm) of \(10\) students in our class \(= 130, 132, 135, 137, 139, 140, 142, 143, 145, 148\)
By observing the above-mentioned values, the highest value is \(= 148\) cm
By observing the above-mentioned values, the lowest value is \(= 130\) cm
Range of Heights = Highest value – Lowest value \(= 148 – 130 = 18\) cm
2. Organise the following marks in a class assessment in a tabular form.
\(4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7\)
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Solution:-
Solutions:-
\(58, 76, 40, 35, 46, 45, 0, 100\).
Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Total number of innings = 8
Mean= \(\dfrac{400}{8}= 50\)
∴ The mean score of the cricketer is \(= 50\).
\(4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7\)
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Solution:-
Arrange the given marks in ascending order \(= 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9\)
Frequency table of the given data:
Frequency table of the given data:
(i)The highest number among the given data is \(9\).
(ii) The lowest number among the given data is \(1\).
(iii) We know that \(\text{Range = Highest value – Lowest value}= 9 – 1= 8\)
(iv) \(\text{Arithmetic mean} = \dfrac{\text{Sum of all observations}}{\text{Total number of observations}}\)
Sum of all observations = 1+2+2+3+4+4+4+5+5+5+5+5+6+6+6+6+7+7+8+9=100
Total number of observations \(= 20\)
Arithmetic mean = \(\dfrac{100}{20}=5\)
Solutions:-
The first five whole numbers are \(0, 1, 2, 3\) and \(4\).
\(\text{Mean} = \dfrac{\text{(Sum of first five whole numbers)}}{ \text{(Total number of whole numbers)}}\)
Sum of five whole numbers \(= 0 + 1 + 2 + 3 +4 = 10\)
Total number of whole numbers \(= 5\)
Mean =\( \dfrac{10}{5}=2\)
∴ The mean of the first five whole numbers is \(2\).
4. A cricketer scores the following runs in eight innings:\(58, 76, 40, 35, 46, 45, 0, 100\).
Find the mean score.
Solution:-
\(\text{Mean score} = \dfrac{\text{Total runs scored by the cricketer in all innings}}{\text{Total number of innings played by the cricketer}}\)
Solution:-
\(\text{Mean score} = \dfrac{\text{Total runs scored by the cricketer in all innings}}{\text{Total number of innings played by the cricketer}}\)
Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Total number of innings = 8
Mean= \(\dfrac{400}{8}= 50\)
∴ The mean score of the cricketer is \(= 50\).
5. Following table shows the points each player scored in four games:
(i) A’s average number of points scored per game = \(\dfrac{\text{Total points scored by A in 4 games}}{\text{Total number of games}}\)
=\( \dfrac{14 + 16 + 10 + 10}{ 4}\)
=\( \dfrac{50}{4}\)
= \(12.5\) points
(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.
(iii) B played in all four games, so we will divide the total points by 4 to find out the mean.
Mean of B’s score = \(\text{Total points scored by B in 4 games}\over \text{Total number of games}\)
=\( \dfrac{(0 + 8 + 6 + 4)}{4}\)
=\(\dfrac{18}{4}\)
= 4.5 points
Answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?
(i) A’s average number of points scored per game = \(\dfrac{\text{Total points scored by A in 4 games}}{\text{Total number of games}}\)
=\( \dfrac{14 + 16 + 10 + 10}{ 4}\)
=\( \dfrac{50}{4}\)
= \(12.5\) points
(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.
(iii) B played in all four games, so we will divide the total points by 4 to find out the mean.
Mean of B’s score = \(\text{Total points scored by B in 4 games}\over \text{Total number of games}\)
=\( \dfrac{(0 + 8 + 6 + 4)}{4}\)
=\(\dfrac{18}{4}\)
= 4.5 points
(iv) Now, we have to find the best performer among the 3 players.
So, we have to find the average points of C= \( \dfrac{8 + 11 + 13}{3}= \dfrac{32}{3}\)
= 10.67 points
By observing, the average points scored by A is 12.5, which is more than by B and C.
Therefore, we can say that A is the best performer among the three.
So, we have to find the average points of C= \( \dfrac{8 + 11 + 13}{3}= \dfrac{32}{3}\)
= 10.67 points
By observing, the average points scored by A is 12.5, which is more than by B and C.
Therefore, we can say that A is the best performer among the three.
6. The marks (out of 100) obtained by a group of students in a science test are \(85, 76,
90, 85, 39, 48, 56, 95, 81\) and \(75\). Find the following:
(i) The highest and lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:-
90, 85, 39, 48, 56, 95, 81\) and \(75\). Find the following:
(i) The highest and lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:-
First, arrange the marks in ascending order, \(= 39, 48, 56, 75, 76, 81, 85, 85, 90, 95\)
(i) The highest marks obtained by the student \(= 95\)
The lowest marks obtained by the student \(= 39\)
(ii) \(\text{ Range = Highest marks – Lowest marks}\)
\(= 95 – 39\)
\(= 56\)
(iii) \(\text{Mean of Marks} = \dfrac{\text{Sum of all marks obtained by the group of students}}{ \text{Total number of marks}}\)
=\( \dfrac{39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95}{ 10}\)
=\( \dfrac{730}{10}\)
= \(73\)
(i) The highest marks obtained by the student \(= 95\)
The lowest marks obtained by the student \(= 39\)
(ii) \(\text{ Range = Highest marks – Lowest marks}\)
\(= 95 – 39\)
\(= 56\)
(iii) \(\text{Mean of Marks} = \dfrac{\text{Sum of all marks obtained by the group of students}}{ \text{Total number of marks}}\)
=\( \dfrac{39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95}{ 10}\)
=\( \dfrac{730}{10}\)
= \(73\)
7. The enrolment in a school for six consecutive years was as follows:
\(1555, 1670, 1750, 2013, 2540, 2820\).
Find the mean enrolment of the school for this period.
Solution:-
Mean enrolment = \(\text{Sum of all observations}\over \text{Number of observations}\)
=\(\dfrac{1555 + 1670 + 1750 + 2013 + 2540 + 2820}{ 6} \)
=\(\dfrac{12348}{6})\)
=\( 2058\)
∴ The mean enrolment of the school for this given period is \(2058\).
8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
\(1555, 1670, 1750, 2013, 2540, 2820\).
Find the mean enrolment of the school for this period.
Solution:-
=\(\dfrac{1555 + 1670 + 1750 + 2013 + 2540 + 2820}{ 6} \)
=\(\dfrac{12348}{6})\)
=\( 2058\)
∴ The mean enrolment of the school for this given period is \(2058\).
8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
(i) \(\text{Range of rainfall = Highest rainfall – Lowest rainfall}\)
\(= 20.5 – 0.0\)
\(= 20.5\) mm
(ii) \(\text{Mean of rainfall} = \dfrac{\text{Sum of all observations}}{\text{Number of observation}}\)
= \(\dfrac{(0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)}{7}\)
= \(\dfrac{41.3}{7}\)
= \(5.9\) mm
(iii) We may observe that for 5 days, i.e., Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.
9. The heights of \(10\) girls were measured in cm, and the results are as follows:
\(135, 150, 139, 128, 151, 132, 146, 149, 143, 141\).
(i) What is the height of the tallest girl?
\(= 20.5 – 0.0\)
\(= 20.5\) mm
(ii) \(\text{Mean of rainfall} = \dfrac{\text{Sum of all observations}}{\text{Number of observation}}\)
= \(\dfrac{(0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)}{7}\)
= \(\dfrac{41.3}{7}\)
= \(5.9\) mm
(iii) We may observe that for 5 days, i.e., Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.
9. The heights of \(10\) girls were measured in cm, and the results are as follows:
\(135, 150, 139, 128, 151, 132, 146, 149, 143, 141\).
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?
Solution:-
First, we have to arrange the given data in ascending order,
\(= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151\)
(i) The height of the tallest girl is \(151\) cm
(ii) The height of the shortest girl is \(128\) cm
(iii) Range of given data = Tallest height – Shortest height \(= 151 – 128= 23\) cm
(iv) Mean height of the girls = \(\dfrac{\text{Sum of the height of all the girls}}{\text{Number of girls}}\)
=\(\dfrac{128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151}{ 10}\)
= \(\dfrac{1414}{10}\)
= \(141.4\) cm
(v) \(5\) girls have heights more than the mean height (i.e. \(141.4\) cm).
(v) How many girls have heights more than the mean height?
Solution:-
\(= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151\)
(i) The height of the tallest girl is \(151\) cm
(ii) The height of the shortest girl is \(128\) cm
(iii) Range of given data = Tallest height – Shortest height \(= 151 – 128= 23\) cm
(iv) Mean height of the girls = \(\dfrac{\text{Sum of the height of all the girls}}{\text{Number of girls}}\)
=\(\dfrac{128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151}{ 10}\)
= \(\dfrac{1414}{10}\)
= \(141.4\) cm
(v) \(5\) girls have heights more than the mean height (i.e. \(141.4\) cm).
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