### Question 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:**

The greatest frequency in the given table is 23, so the modal class = 35 â€“ 45,

l = 35,

Class width = 10, and the frequencies are

f

_{m}= 23, f_{1}= 21 and f_{2}= 14Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

x

_{i }= (upper limit + lower limit)/2

Class IntervalFrequency (f_{i})Mid-point (x_{i})f_{i}x_{i}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 Sum f_{i}= 80Sum f_{i}x_{i}= 2830Mean = = âˆ‘f

_{i}x_{i}/âˆ‘f_{i}= 2830/80

= 35.37 years

### Question 2. The following data gives the information on the observed lifetimes (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:**

According to the given question

The modal class is 60 â€“ 80

l = 60, and the frequencies are

f

_{m}= 61, f_{1}= 52, f_{2}= 38 and h = 20Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

=

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 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:

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:**

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

f

_{m}= 40 f_{1}= 24, f_{2}= 33 andh = 500

Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

=

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees

Now, we find the mean. So for that first we need to find the midpoint.

x

_{i }= (upper limit + lower limit)/2Let us considered a mean, A be 2750

Class Intervalf_{i}x_{i}d_{i}= x_{i}â€“ au_{i}= d_{i}/hf_{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 f_{i}= 200f_{i}u_{i}= -35Mean =

On substituting the values in the given formula

=

= 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is 2662.50 Rupees

### 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

No of Students per teacher | Number of states / U.T |

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:**

According to the question

Modal class = 30 â€“ 35,

l= 30,Class width (h) = 5, and the frequencies are

f

_{m}= 10, f_{1}= 9 and f_{2}= 3Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

x

_{i }= (upper limit + lower limit)/2

Class IntervalFrequency (f_{i})Mid-point (x_{i})f_{i}x_{i}15-20 3 17.5 52.5 20-25 8 22.5 180.0 25-30 9 27.5 247.5 30-35 10 32.5 325.0 35-40 3 37.5 112.5 40-45 0 42.5 0 45-50 0 47.5 0 50-55 2 52.5 105.5 Sum f_{i}= 35Sum f_{i}x_{i}= 1022.5Mean =

= 1022.5/35

= 29.2

Hence, the mean is 29.2

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

Run Scored | Number of Batsman |

3000-4000 | 4 |

4000-5000 | 18 |

5000-6000 | 9 |

6000-7000 | 7 |

7000-8000 | 6 |

8000-9000 | 3 |

9000-10000 | 1 |

10000-11000 | 1 |

### Find the mode of the data.

**Solution:**

According to the question

Modal class = 4000 â€“ 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

f

_{m}= 18, f_{1}= 4 and f_{2}= 9Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given 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 summarized it in the table given below. Find the mode of the data:

Number of cars | Frequency |

0-10 | 7 |

10-20 | 14 |

20-30 | 13 |

30-40 | 12 |

40-50 | 20 |

50-60 | 11 |

60-70 | 15 |

70-80 | 8 |

**Solution:**

According to the question

Modal class = 40 â€“ 50, l = 40,

Class width (h) = 10, and the frequencies are

f

_{m}= 20, f_{1}= 12 and f_{2}= 11Now, we find the mode using the given formula

Mode

=On substituting the values in the formula, we get

Mode =

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars