**Find the range****of heights of any ten students of your class.**

**Solution:**

Let us taken the height of 10 students in cm

150, 122, 135, 166, 117, 118, 113, 100, 112, 132

Now, By observing this data we find the highest value =166

By observing this data we find the lowest value = 100

Then,

Range of heights = Highest value – Lowest value

= 166 – 100

= 66 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**

**Note:** First we have to arrange the given marks in ascending order:

i.e. 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9

Let us form a frequency table of the given data.

Marks (x_{i}) | Tally Marks | Frequency (f_{i}) | f_{i} x_{i} |

1 | I | 1 | 1 |

2 | II | 2 | 4 |

3 | I | 1 | 3 |

4 | III | 3 | 12 |

5 | 5 | 25 | |

6 | IIII | 4 | 24 |

7 | II | 2 | 14 |

8 | I | 1 | 8 |

9 | I | 1 | 9 |

Total – | 20 | 100 |

**(i) Which number is the highest? **

**Solution:**

By observing the given data, the highest number is = 9

**(ii) Which number is the lowest?**

**Solution:**

By observing the given data, the lowest number is = 1

**(iii) What is the range of the data? **

**Solution:**

Range = Highest value – Lowest value

= 9 – 1

= 8

Hence, the range of the data is 8.

**(iv) Find the arithmetic mean.**

**Solution:**

= 100/20

= 5

**3. Find the mean of the first five whole numbers.**

**Solutions:**

**4. A cricketer scores the following runs in eight innings:**

**58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.**

**Solution:**

**5. Following table shows the points of each player scored in four games:**

Player | Game 1 | Game 2 | Game 3 | Game 4 |

A | 14 | 16 | 10 | 10 |

B | 0 | 8 | 6 | 4 |

C | 8 | 11 | Did not Play | 13 |

**Now answer the following questions:**

**(i) Find the mean to determine A’s average number of points scored per game.**

**Solution:**

A’s average number of points scored per day = total points scored by a in 4 games/Total number of games

= (14+16+10+10)/4

= 50/4 = 12.5 points

**(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?**

**Solution:**

we divide by 3 to find the mean number of points per game for C because C played only 3 games.

**(iii) B played in all the four games. How would you find the mean?**

**Solution:**

Total number of games = 4

Given, B played in all the four games

So, To find out the mean we will divide the total points by 4

Now,

Mean of B’s score = Total point scored by B in 4 games/Total number of games

= (0+8+6+4)/4

= 18/4 = 4.5 points

**(iv) Who is the best performer?**

**Solution:**

To find the best performer among 3 player

First, we have to find the average point of C = (8+11+13)/3

= 32/3 = 10.66

Now, we have average points scored by A is 12.5 which is more than B and C.

Hence, it is clear that A is the best performer among the 3 player.

**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:**

First of all, we have to arrange the marks in ascending order:

= 39,48,56,75,76,81,85,85,90,95

**(i) Highest and the lowest marks obtained by the students.**

**Solution:**

By observing the given marks:

The highest marks obtained by the student = 95

The lowest marks obtained by the student=39

**(ii) Range of the marks obtained.**

**Solution:**

Range of marks = Highest marks – Lowest marks

= 95-39 = 56

**(iii) Mean marks obtained by the group.**

**Solution:**

Mean of marks obtained by group = (sum of all marks obtained by the group of students)/(Total number of marks)

= (39+48+56+75+76+81+85+85+90+95)/10

= 730/10 = 73

**7. The enrolment in a school during 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 = (Sum of all observation )/ (Number of Observation)

= ( 1555 + 1670 + 1750 + 2013 + 2540 + 2820)/6

= 12348/6 = 2058

Hence, the required mean enrolment of a school for the given period is 2058

**8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:**

Day | Mon | Tue | Wed | Thurs | Fri | Sat | Sun |

Rainfall (in mm) | 0.0 | 12.2 | 2.1 | 0.0 | 20.5 | 5.5 | 1.0 |

**(i) Find the range of the rainfall in the above data.**

**Solution:**

Range of rainfall for the given data = highest rainfall – lowest rainfall

= 20.5 – 0.0 = 20.5 mm

**(ii) Find the mean rainfall for the week.**

**Solution:**

Mean of rainfall for the given data = (Sum of all observation) / (Total number of observation)

= (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0) / 7

= 41.3 / 7 = 5.9 mm

**(iii) On how many days was the rainfall less than the mean rainfall.**

**Solution:**

After observing the average rainfall,

We may find the rainfall of 5 days was less than the average rainfall i.e Monday, Wednesday, Thursday, Saturday, and Sunday.

**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.**

**We have to arrange the given data in ascending order:= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151**

**(i) What is the height of the tallest girl? **

**Solution:**

By observing the given data the height of the tallest girl = 151 cm

**(ii) What is the height of the shortest girl?**

**Solution:**

By observing the given data the height of the shortest girl = 128 cm

**(iii) What is the range of the data? **

**Solution:**

Range of the given data = Tallest height – Shortest height

= 151 – 128 = 23 cm

**(iv) What is the mean height of the girls?**

**Solution:**

Mean height of the girl = (Sum of the height of all-girl) / Total number of girls)

= (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151) / 10

= 1414 / 10 = 141.4 cm

**(v) How many girls have heights more than the mean height.**

**Solution:**

The mean height is 141.4 cm

Hence, 5 girls have a height more than the mean height.

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