# Ex.14.3 Q4 Statistics Solution - NCERT Maths Class 9

## Question

The length of \(40\) leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm) |
Number of leaves |

118– 126 | 3 |

127– 135 | 5 |

136– 144 | 9 |

145– 153 | 12 |

154– 162 | 5 |

163– 171 | 4 |

172- 180 | 2 |

- Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
- Is there any other suitable graphical representation for the same data?
- Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

**Difficulty Level: Medium ** **Reasoning:**

- From the given data, it can be observed that the length of leaves are represented in discontinuous class intervals, with a difference of 1 unit in between them.
- To make the class intervals continuous, we can add 0.5

**What is known/given?**

The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the table.

**What is unknown? **

- A histogram to represent the given data
- Class interval type
- Another way of graphical representation of the same data.
- As certaining the conclusion that the maximum number of leaves are 153 mm long.

**Solution:**

Find the difference between upper limit of a class and the lower limit of its succeeding class. We then add half of this difference to each of the upper limits and subtract the same from each of the lower limit since the difference is 1 (127 – 126 = 1) half of 1 is 0.5. Now the frequency distribution table will look like as below.

Length (in mm) |
Number of Leaves |

117.5 – 126.5 | 3 |

126.5 – 135.5 | 5 |

135.5 – 144.5 | 9 |

144.5 – 153.5 | 12 |

153.5 – 162.5 | 5 |

162.5 – 171.5 | 4 |

171.5 – 180.5 | 2 |

The above data can be represented graphically through a histogram as below:

- Represent “Length of leaves (in mm) on x-axis and ‘number of leaves’ in y-axis.
- And with a scale of ‘1 unit = 2 leaves’, since the lower-class value is 2 and the highest is 12.
- Also, since the first-class interval is starting from 117.5 and not zero, we show it on the graph by making a kink on a break on the x-axis.
- We will make now draw rectangular bars of width equal in sizes and lengths according to the frequencies of class intervals. For eg, the rectangular bar for the class interval 117.5 – 126.5 will be of width 1cm and length 1.5 cm.

<Insert image (14.3-5)> It can be observed from the above graph that the other suitable graphical representation of the same data would be a frequency polygon as below: Join the mid points of the upper sides of all the rectangular bars through line segments (as shown in red dotted lines) Let them B, C, D, E, F, G, H. To complete it look like a polygon figure, assume that there is a class interval with ‘0’ frequency, before ‘117.5 – 126.5’, and one after ‘171.5 – 180.5’. Now mark their imaginary mid points as ‘a’ and ‘I’ respectively. Now, “ABCDEFGHI” is the frequency polygon that can be constructed for the given data. (Refer the red dotted line segment) The maximum number of leaves have their length lie between 144.5 mm and 153.5 mm. Hence, we cannot say all leaves have their lengths as 153 mm.