Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units) | Number of customers |
---|---|

65-85 | 4 |

85-105 | 5 |

105-125 | 13 |

125-145 | 20 |

145-165 | 14 |

165-185 | 8 |

185-205 | 4 |

**Solution:**

Class Interval | Frequency | Cumulative frequency |
---|---|---|

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 8 | 64 |

185-205 | 4 | 68 |

N = 68 |

Here; n = 68 and hence `n/2 = 34`

So, median class is 125-145 with cumulative frequency = 42

now, `l = 125`, `n = 68`, `cf = 22`, `f = 20`, `h = 20`

Median can be calculated as follows:

`text(Median)=l+(n/2-cf)/(f)xxh`

`=125+(34-22)/(20)xx20`

`=125+12=137`

Modal class `= 125-145`, `f_1 = 20`, `f_0 = 13`, `f_2 = 14` and `h = 20`

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

`=125+(20-13)/(2xx20-13-14)xx20`

`=125+7/13xx20`

`=125+10.77=135.77`

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

65-85 | 4 | 75 | -60 | -3 | -12 |

85-105 | 5 | 95 | -40 | -2 | -10 |

105-125 | 13 | 115 | -20 | -1 | -13 |

125-145 | 20 | 135 | 0 | 0 | 0 |

145-165 | 14 | 155 | 20 | 1 | 14 |

165-185 | 8 | 175 | 40 | 2 | 16 |

185-205 | 4 | 195 | 60 | 3 | 12 |

Σ f_{i} = 68 | Σ f_{i}u_{i} = 7 |

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=135+7/68xx20=137.05`

Mean, median and mode are more or less equal in this distribution.

Question 2. If the median of the distribution given below is 28.5, find the value of x and y.

Class Interval | Frequency |
---|---|

0-10 | 5 |

10-20 | x |

20-30 | 20 |

30-40 | 15 |

40-50 | y |

50-60 | 5 |

Total | 60 |

**Solution:** `n = 60` and hence `n/2 = 30`

Median class is 20 – 30 with cumulative frequency `= 25 + x`

lower limit of median class = 20, `cf = 5 + x`, `f = 20` and `h = 10`

`text(Median)=l+(n/2-cf)/(f)xxh`

Or, `28.5=20+(30-5-x)/(20)xx10`

Or, `(25-x)/(2)=8.5`

Or, `25-x=17`

Or, `x=25-17=18`

Now, from cumulative frequency, we can find the value of x + y as follows:

`60=5+20+15+5+x+y`

Or, `45+x+y=60`

Or, `x+y=60-45=15`

Hence, `y=15-x=15-8=7`

Hence, x = 8 and y = 7

Question 3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years) | Number of policy holders |
---|---|

Below 20 | 2 |

Below 25 | 6 |

Below 30 | 24 |

Below 35 | 45 |

Below 40 | 78 |

Below 45 | 89 |

Below 50 | 92 |

Below 55 | 98 |

Below 60 | 100 |

**Solution:**

Class interval | Frequency | Cumulative frequency |
---|---|---|

15-20 | 2 | 2 |

20-25 | 4 | 6 |

25-30 | 18 | 24 |

30-35 | 21 | 45 |

35-40 | 33 | 78 |

40-45 | 11 | 89 |

45-50 | 3 | 92 |

50-55 | 6 | 98 |

55-60 | 2 | 100 |

Here; `n = 100` and `n/2 = 50`, hence median class `= 35-45`

In this case; `l = 35`, `cf = 45`, `f = 33` and `h = 5`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=35+(50-45)/(33)xx5`

`=35+25/33=35.75`

Question 4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained 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 |

Find the median length of leaves.

**Solution:**

Class Interval | Frequency | Cumulative frequency |
---|---|---|

117.5-126.5 | 3 | 3 |

126.5-135.5 | 5 | 8 |

135.5-144.5 | 9 | 17 |

144.5-153.5 | 12 | 29 |

153.5-162.5 | 5 | 34 |

162.5-171.5 | 4 | 38 |

171.5-180.5 | 2 | 40 |

We have; `n = 40` and `n/2 = 20` hence median class `= 144.5-153.5`

Thus, `l = 144.5`, `cf = 17`, `f = 12` and `h = 9`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=144.5+(20-17)/(12)xx9`

`=144.5+9/4=146.75`

Question 5. The following table gives distribution of the life time of 400 neon lamps.

Lifetime (in hours) | Number of lamps |
---|---|

1500-2000 | 14 |

2000-2500 | 56 |

2500-3000 | 60 |

3000-3500 | 86 |

3500-4000 | 74 |

4000-4500 | 62 |

4500-5000 | 48 |

Find the median life time of a lamp.

**Solution:**

Class Interval | Frequency | Cumulative Frequency |
---|---|---|

1500-2000 | 14 | 14 |

2000-2500 | 56 | 70 |

2500-3000 | 60 | 130 |

3000-3500 | 86 | 216 |

3500-4000 | 74 | 290 |

4000-4500 | 62 | 352 |

4500-5000 | 48 | 400 |

We have; `n = 400` and `n/2 = 200` hence median class `= 3000 – 3500`

So, `l = 3000`, `cf = 130`, `f = 86` and `h = 500`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=3000+(200-130)/(86)xx500`

`=3000+406.97=3406.97`

Question 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |
---|---|---|---|---|---|---|

Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

**Solution:** Calculations for median:

Class Interval | Frequency | Cumulative Frequency |
---|---|---|

1-4 | 6 | 6 |

4-7 | 30 | 36 |

7-10 | 40 | 76 |

10-13 | 16 | 92 |

13-16 | 4 | 96 |

16-19 | 4 | 100 |

Here; `n = 100` and `n/2 = 50` hence median class `= 7-10`

So, `l = 7`, `cf = 36`, `f = 40` and `h = 3`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=7+(50-36)/(40)xx3`

`=7+14/40xx3=8.05`

Modal class `= 7-10`

Here; `l = 7`, `f_1 = 40`, `f_0 = 30`, `f_2 = 16` and `h = 3`

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

`=7+(40-30)/(2xx40-30-16)xx3`

`=7+10/34xx3=7.88`

Class interval | fi | xi | fixi |
---|---|---|---|

1-4 | 6 | 2.5 | 15 |

4-7 | 30 | 5.5 | 165 |

7-10 | 40 | 8.5 | 340 |

10-13 | 16 | 11.5 | 184 |

13-16 | 4 | 14.5 | 51 |

16-19 | 4 | 17.5 | 70 |

Σ f_{i} = 100 | Σ f_{i}x_{i} = 825 |

`x= (Σf_i\x_i)/(Σf_i)`

`=(825)/(100)=8.25`

Question 7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
---|---|---|---|---|---|---|---|

Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |

**Solution:**

Class Interval | Frequency | Cumulative frequency |
---|---|---|

40-45 | 2 | 2 |

45-50 | 3 | 5 |

50-55 | 8 | 13 |

55-60 | 6 | 19 |

60-65 | 6 | 25 |

65-70 | 3 | 28 |

70-75 | 2 | 30 |

We have; `n = 30` and `n/2 = 15` hence median class `= 55-60`

So, `l = 55`, `cf = 13`, `f = 6` and `h = 5`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=55+(15-13)/(6)xx5`

`=55+5/3=56.67`

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