__Exercise 14.2 Page: 275__

__Exercise 14.2 Page: 275__

**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 twomeasures of central tendency.**

Solution:

To find out the modal class, let us the consider the class interval with high frequency

Here, the greatest frequency = 23, so the modal class = 35 – 45,

l = 35,

class width (h) = 10,

f_{m} = 23,

f_{1} = 21 and f_{2} = 14

The formula to find the mode is

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

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

Mode = 36.8 year

So the mode of the given data = 36.8 year

Calculation of Mean:

First find the midpoint using the formula, x_{i }= (upper limit +lower limit)/2

Class Interval | Frequency (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} = 80 |
Sum f_{i}x_{i} = 2830 |

The mean formula is

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 years

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

From the given data the modal class is 60–80.

l = 60,

The frequencies are:

f_{m} = 61, f_{1} = 52, f_{2} = 38 and h = 20

The formula to find the mode is

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 20)/32)

Mode = 60+(45/8) = 60+ 5.625

Therefore, modal lifetime of the components = 65.625 hours.

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

Given data:

Modal class = 1500-2000,

*l* = 1500,

Frequencies:

f_{m} = 40 f_{1} = 24, f_{2} = 33 and

h = 500

Mode formula:

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values in the formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/23)

Mode = 1500+(8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, x_{i }=(upper limit +lower limit)/2

Let us assume a mean, A be 2750

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 |

fi = 200 | fiui = -35 |

The formula to calculate the mean,

Mean = x̄ = a +(∑f_{i}u_{i} /∑f_{i})×h

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 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**

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:

Given data:

Modal class = 30 – 35,

*l* = 30,

Class width (h) = 5,

f_{m} = 10, f_{1} = 9 and f_{2} = 3

Mode Formula:

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x_{i }=(upper limit +lower limit)/2

Class Interval | Frequency (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} = 35 |
Sum f_{i}x_{i} = 1022.5 |

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}

= 1022.5/35

= 29.2

Therefore, mean = 29.2

**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:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

f_{m} = 18, f_{1} = 4 and f_{2} = 9

Mode Formula:

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs

**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:

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, f_{m} = 20, f_{1} = 12 and f_{2} = 11

Mode *= l+ [(f _{m}-f_{1})/(2f_{m}-f_{1}-f_{2})]×h*

Substitute the values

Mode = 40+((20-12)/(40-12-11))×10

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

Thus, the mode of the given data is 44.7 cars