TOPICS

Exercise - 14.2

Statistics

**Question-1 :-** The following table shows the ages of the patients admitted in a hospital during a year:
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 10 Taking assumed mean (a) = 30

Age (in years) | Number of patients (f_{i}) | x_{i} | d_{i} = x_{i} - 30 | f_{i}d_{i} |

5 - 15 | 6 | 10 | -20 | -120 |

15 - 25 | 11 | 20 | -10 | -110 |

25 - 35 | 21 | 30 | 0 | 0 |

35 - 45 | 23 | 40 | 10 | 230 |

45 - 55 | 14 | 50 | 20 | 280 |

55 - 65 | 5 | 60 | 30 | 150 |

Total | Σf_{i} = 80 | Σf_{i}d_{i} = 430 |

Here the maximum class frequency is 23, and the class corresponding to this frequency is 35 – 45. So, the modal class is 35 – 45. Now, Modal class = 35 - 45, lower limit(l) of modal class = 23, class size(h) = 10 frequency(f_{1}) of modal class = 23, frequency(f_{0}) of modal class = 21, frequency(f_{2}) of modal class = 14, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 35 + [(23-21)/{(2x23)-21-14}] x 10 = 35 + [2/(46-35)] x 10 = 35 + (2/11 x 10) = 35 + 20/11 = 35 + 1.81 = 36.81 Therefore, Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.81 years.

**Question-2 :-** The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Determine the modal lifetimes of the components.

Here the maximum class frequency is 61, and the class corresponding to this frequency is 60 – 80. So, the modal class is 60 – 80. Now, Modal class = 60 - 80, lower limit(l) of modal class = 60, class size(h) = 20 frequency(f_{1}) of modal class = 61, frequency(f_{0}) of modal class = 52, frequency(f_{2}) of modal class = 38, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 60 + [(61-52)/{(2x61)-52-38}] x 20 = 60 + [9/(122-90)] x 20 = 60 + (9/32 x 20) = 60 + 45/8 = 60 + 5.625 = 65.625 = 65.63 (approx.) Therefore, modal lifetime of electrical coponents is 65.63 hours.

**Question-3 :-** The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Here the maximum class frequency is 40, and the class corresponding to this frequency is 1500 – 2000. So, the modal class is 1500 – 2000. Now, Modal class = 1500 - 2000, lower limit(l) of modal class = 1500, class size(h) = 500 frequency(f_{1}) of modal class = 40, frequency(f_{0}) of modal class = 24, frequency(f_{2}) of modal class = 33, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 1500 + [(40-24)/{(2x40)-24-33}] x 500 = 1500 + [16/(80-57)] x 500 = 1500 + (16/23 x 500) = 1500 + 8000/23 = 1500 + 347.826 = 1847.826 = 1847.83 (approx.) Therefore, modal monthly expenditure was Rs. 1847.83.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 500 Taking assumed mean (a) = 2750

Expenditure (in Rs.) | Number of families (f_{i}) | x_{i} | d_{i} = x_{i} - 2750 | u_{i} = d_{i}/500 | f_{i}u_{i} |

1000 - 1500 | 24 | 1250 | -1500 | -3 | -72 |

1500 - 2000 | 40 | 1750 | -1000 | -2 | -80 |

2000 - 2500 | 33 | 2250 | -500 | -1 | -33 |

2500 - 3000 | 28 | 2750 | 0 | 0 | 0 |

3000 - 3500 | 30 | 3250 | 500 | 1 | 30 |

3500 - 4000 | 22 | 3750 | 1000 | 2 | 44 |

4000 - 4500 | 16 | 4250 | 1500 | 3 | 48 |

4500 - 5000 | 7 | 4750 | 2000 | 4 | 28 |

Total | Σf_{i} = 200 | Σf_{i}u_{i} = -35 |

**Question-4 :-** The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Here the maximum class frequency is 10, and the class corresponding to this frequency is 30 – 35. So, the modal class is 30 – 35. Now, Modal class = 30 - 35, lower limit(l) of modal class = 30, class size(h) = 5 frequency(f_{1}) of modal class = 10, frequency(f_{0}) of modal class = 9, frequency(f_{2}) of modal class = 3, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 30 + [(10-9)/{(2x10)-9-3}] x 5 = 30 + [1/(20-12)] x 5 = 30 + (1/8 x 5) = 30 + 5/8 = 30 + 0.625 = 30.625 = 30.63 (approx.) Therefore, the mode of this data is 30.63. It represents that most of the states/U.T have a teacher-student ratio as 30.63.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 5 Taking assumed mean (a) = 32.5

Expenditure (in Rs.) | Number of families (f_{i}) | x_{i} | d_{i} = x_{i} - 32.5 | u_{i} = d_{i}/5 | f_{i}u_{i} |

15 - 20 | 3 | 17.5 | -15 | -3 | -9 |

20 - 25 | 8 | 22.5 | -10 | -2 | -16 |

25 - 30 | 9 | 27.5 | -5 | -1 | -9 |

30 - 35 | 10 | 32.5 | 0 | 0 | 0 |

35 - 40 | 3 | 37.5 | 5 | 1 | 3 |

40 - 45 | 0 | 42.5 | 10 | 2 | 0 |

45 - 50 | 0 | 47.5 | 15 | 3 | 0 |

50 - 55 | 2 | 52.5 | 20 | 4 | 8 |

Total | Σf_{i} = 35 | Σf_{i}u_{i} = -23 |

**Question-5 :-** The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Find the mode of the data.

Here the maximum class frequency is 18, and the class corresponding to this frequency is 4000 – 5000. So, the modal class is 4000 – 5000. Now, Modal class = 4000 - 5000, lower limit(l) of modal class = 4000, class size(h) = 1000 frequency(f_{1}) of modal class = 18, frequency(f_{0}) of modal class = 4, frequency(f_{2}) of modal class = 9, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 4000 + [(18-4)/{(2x18)-4-9}] x 1000 = 4000 + [14/(36-13)] x 1000 = 4000 + (14/23 x 1000) = 4000 + 14000/23 = 4000 + 608.695 = 4608.695 = 4608.7 (approx.) Therefore, the mode of this data is 4608.7 runs.

**Question-6 :-** A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Here the maximum class frequency is 20, and the class corresponding to this frequency is 40 – 50. So, the modal class is 40 – 50. Now, Modal class = 40 - 50, lower limit(l) of modal class = 40, class size(h) = 10 frequency(f_{1}) of modal class = 20, frequency(f_{0}) of modal class = 12, frequency(f_{2}) of modal class = 11, Therefore, Mode = l + [(f_{1}- f_{0})/(2f_{1}- f_{0}- f_{2})] x h = 40 + [(20-12)/{(2x20)-12-11}] x 10 = 40 + [8/(40-23)] x 10 = 40 + (8/17 x 10) = 40 + 80/17 = 40 + 4.7 = 44.7 Therefore, the mode of this data is 44.7 cars.

CLASSES