1. The following table shows the ages of the patients admitted in a hospital during a year.

Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
---|---|---|---|---|---|---|

Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

**Solution:** Solution: Modal class `= 35 – 45`, `l = 35`, `h = 10`, `f_1 = 23`, `f_0 = 21` and `f_2 = 14`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=35+(23-21)/(2xx23-21-14)xx10`

`=35+2/11xx10=36.8`

Class Interval | fi | xi | fixi |
---|---|---|---|

5-15 | 6 | 10 | 60 |

15-25 | 11 | 20 | 220 |

25-35 | 21 | 30 | 630 |

35-45 | 23 | 40 | 920 |

45-55 | 14 | 50 | 700 |

55-65 | 5 | 60 | 300 |

Σ f_{i} = 80 | Σ f_{i}x_{i} = 2830 |

`x=(Σf_i\x_i)/(Σf_i)=(2830)/(80)=35.37`

The mode of the data shows that maximum number of patients is in the age group of 26.8, while average age of all the patients is 35.37.

2. The following data gives the information on the observed lifetime (in hours) of 225 electrical components:

Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
---|---|---|---|---|---|---|

Frequency | 10 | 35 | 52 | 61 | 38 | 29 |

Determine the modal lifetimes of the components.

**Solution:** Modal class `= 60-80`, `l = 60`, `f_1 = 61`, `f_0 = 52`, `f_2 = 38` and `h = 20`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=60+(61-52)/(2xx61-52-38)xx20`

`=60+9/32xx20=65.62`

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.

Expenditure | Number of families |
---|---|

1000-1500 | 24 |

1500-2000 | 40 |

2000-2500 | 33 |

2500-3000 | 28 |

3000-3500 | 30 |

3500-4000 | 22 |

4000-4500 | 16 |

4500-5000 | 7 |

**Solution:** Modal class `= 1500-2000`, `l = 1500`, `f_1 = 40`, `f_0 = 24`, `f_2 = 33` and `h = 500`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=1500+(40-24)/(2xx40-24-33)xx500`

`=1500+16/23xx500=1847.82`

Class Interval | fi | xi | di = xi - a | ui = di/h | fiui |
---|---|---|---|---|---|

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 |

Σ f_{i} = 200 | Σ f_{i}u_{i} = -35 |

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=2750-(35)/(200)xx500`

`=2750-87.5=2662.50`

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.

Number of students per teacher | Number of states/UT |
---|---|

15-20 | 3 |

20-25 | 8 |

25-30 | 9 |

30-35 | 10 |

35-40 | 3 |

40-45 | 0 |

45-50 | 0 |

50-55 | 2 |

**Solution:** Modal class `= 30-35`, `l = 30`, `f_1 = 10`, `f_0 = 9`, `f_2 = 3` and `h = 5`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=30+(10-9)/(2xx10-9-3)xx5`

`=30+1/8xx5=30.625`

Class Interval | fi | xi | di = xi - a | ui = di/h | fiui |
---|---|---|---|---|---|

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 |

Σ f_{i} = 35 | Σ f_{i}u_{i} = -23 |

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=32.5-23/35xx5`

`=32.5-22/7=29.22`

The mode shows that maximum number of states has 30-35 students per teacher. The mean shows that average ratio of students per teacher is 29.22

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored | Number of batsmen |
---|---|

3000-4000 | 4 |

4000-5000 | 18 |

5000-6000 | 9 |

6000-7000 | 7 |

7000-8000 | 6 |

8000-9000 | 3 |

9000-1000 | 1 |

10000-11000 | 1 |

Find the mode of the data.

**Solution:** Modal class `= 4000-5000`, `l = 4000`, `f_1 = 18`, `f_0 = 4`, `f_2 = 9` and `h = 1000`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=4000+(18-4)/(2xx18-4-9)xx1000`

`=4000+14/23xx1000=4608.70`

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

Number of cars | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
---|---|---|---|---|---|---|---|---|

Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |

**Solution:** Modal class `= 40 – 50`, `l = 40`, `f_1 = 20`, `f_0 = 12`, `f_2 = 11` and `h = 10`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=40+(20-12)/(2xx20-12-11)xx10`

`=40+8/17xx10=44.70`

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