### Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Daily income in Rupees | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |

Number of workers | 12 | 14 | 8 | 6 | 10 |

**Solution:**

According to the question, we convert the given distribution to a less than type cumulative frequency distribution,

Daily incomeFrequencyCumulative FrequencyLess than 120 12 12 Less than 140 14 26 Less than 160 8 34 Less than 180 6 40 Less than 200 10 50 Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve.

### Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight in kg | Number of students |

Less than 38 | 0 |

Less than 40 | 3 |

Less than 42 | 5 |

Less than 44 | 9 |

Less than 46 | 14 |

Less than 48 | 28 |

Less than 50 | 32 |

Less than 52 | 35 |

### Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

**Solution:**

According to the give table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.

Now, locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:

Class intervalNumber of students(Frequency)Cumulative FrequencyLess than 38 0 0 Less than 40 3 – 0 = 3 3 Less than 42 5 – 3 = 2 8 Less than 44 9 – 5 = 4 9 Less than 46 14 – 9 = 5 14 Less than 48 28 – 14 = 14 28 Less than 50 32 – 28 = 4 32 Less than 52 35 – 22 = 3 35 The class 46 – 48 has the maximum frequency, hence, this is the modal class

l= 46, h = 2, f_{1 }= 14, f_{0 }= 5 and f_{2}= 4Now we find the mode:

Mode =

On substituting the values in the given formula, we get

= 46 + 0.95 = 46.95

Hence, the mode is verified.

### Question 3. The following tables give the production yield per hectare of wheat of 100 farms of a village.

Production Yield | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |

Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |

### Change the distribution to a more than type distribution and draw its ogive.

**Solution:**

According to the question, we change the distribution to a more than type distribution.

Production Yield (kg/ha)Number of farmsMore than or equal to 50 100 More than or equal to 55 100 – 2 = 98 More than or equal to 60 98 – 8 = 90 More than or equal to 65 90 – 12 = 78 More than or equal to 70 78 – 24 = 54 More than or equal to 75 54 – 38 = 16 Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.