Question 1:

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method/p>

Literacy rate (%) | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |

Number of cities | 4 | 11 | 12 | 9 | 4 |

Solution:

Using Direct method, the given data is shown as follows:

Literacy rate (%) | Number of cities (f_{i}) |
Class mark (x_{i}) |
(f_{i}x_{i}) |

45 – 55 | 4 | 50 | 200 |

55 – 65 | 11 | 60 | 660 |

65 – 75 | 12 | 70 | 840 |

75 – 85 | 9 | 80 | 720 |

85 – 95 | 4 | 90 | 360 |

Total | \(\sum f_{i}=40\) |
\(\sum (f_{i}\times x_{i})=2780\) |

The mean of given data is given by

\(\bar{x}\)

= \(\frac{2780}{40}\)

= 69.5

Thus, the mean literacy rate is 69.5%

Question 2:

Find the mean of the following frequency distribution using step-deviation method.

Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Frequency | 7 | 10 | 15 | 8 | 10 |

Solution:

Let us choose a = 25, h = 10, then d_{i} = x_{i} – 25 and u_{i} = \(\frac{x_{i}-25}{10}\)

Using Step-deviation method, the given data is shown as follows:

Class | Frequency (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 25 |
u_{i} = \(\frac{x_{i}-25}{10}\) |
f_{i}u_{i} |

0 – 10 | 7 | 5 | -20 | -2 | -14 |

10 – 20 | 10 | 15 | -10 | -1 | -10 |

20 – 30 | 15 | 25 | 0 | 0 | 0 |

30 – 40 | 8 | 35 | 10 | 1 | 8 |

40 – 50 | 10 | 45 | 20 | 2 | 20 |

Total | \(\sum f_{i}=50\) |
\(\sum (f_{i}\times u_{i})=4\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 25 + \(\frac{4}{50}\)

= 25 + \(\frac{4}{5}\)

= \(\frac{125+4}{5}\)

= \(\frac{129}{5}\)

= 25.8

Thus, the mean is 25.8.

Question 3:

Find the mean of the following data, using step-deviation method:

Class | 5 – 15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 |

Frequency | 6 | 10 | 16 | 15 | 24 | 8 | 7 |

Solution:

Let us choose a = 40, h = 10, then d_{i} = x_{i} – 40 and u_{i} = \(\frac{x_{i}-40}{10}\)

Using Step-deviation method, the given data is shown as follows:

Class | Frequency (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 40 |
u_{i} = \(\frac{x_{i}-40}{10}\) |
f_{i}u_{i} |

5 – 15 | 6 | 10 | -30 | -3 | -18 |

15 – 25 | 10 | 20 | -20 | -2 | -20 |

25 – 35 | 16 | 30 | -10 | -1 | -16 |

35 – 45 | 15 | 40 | 0 | 0 | 0 |

45 – 55 | 24 | 50 | 10 | 1 | 24 |

55 – 65 | 8 | 60 | 20 | 2 | 16 |

65 – 75 | 7 | 70 | 30 | 3 | 21 |

Total | \(\sum f_{i}=86\) |
\(\sum (f_{i}\times u_{i})=7\) |

The mean of given data is given by/p>

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 40 + \(\frac{7}{86}\)

= 40 + \(\frac{70}{86}\)

= 40 + 0.81

= 40.81

Thus, the mean is 40.81.

Question 4:

The weights of tea in 70 packets are shown in the following table

Weights (in grams) | 200 – 201 | 201 – 202 | 202 – 203 | 203 – 204 | 204 – 205 | 205 – 206 |

Number of packets | 13 | 27 | 18 | 10 | 1 | 1 |

Find the mean weight of packets using step-deviation method.

Solution:

Let us choose a = 202.5, h = 1, then d_{i} = x_{i} – 202.5 and u_{i} = \(\frac{x_{i}-202.5}{1}\)

Using Step-deviation method, the given data is shown as follows:

Weight (in grams) | Number of packets (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 202.5 |
u_{i} = \(\frac{x_{i}-202.5}{1}\) |
f_{i}u_{i} |

200 – 201 | 13 | 200.5 | -2 | -2 | -26 |

201 – 202 | 27 | 201.5 | -1 | -1 | -27 |

202 – 203 | 18 | 202.5 | 0 | 0 | 0 |

203 – 204 | 10 | 203.5 | 1 | 1 | 10 |

204 – 205 | 1 | 204.5 | 2 | 2 | 2 |

205 – 206 | 1 | 205.5 | 3 | 3 | 3 |

Total | \(\sum f_{i}=70\) |
\(\sum (f_{i}\times u_{i})=-38\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 202.5 + \(\frac{-38}{70}\)

= 202.5 – 0.542

= 201.96

Hence, the mean is 201.96 g.

Question 5:

Find the mean of the following frequency distribution using a suitable method.

Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |

Frequency | 25 | 40 | 42 | 33 | 10 |

Solution:

Let us choose a = 45, h = 10, then d_{i} = x_{i} – 45 and u_{i} = \(\frac{x_{i}-45}{10}\)

Using Step-deviation method, the given data is shown as follows:

Class | Frequency (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 45 |
u_{i} = \(\frac{x_{i}-45}{10}\) |
f_{i}u_{i} |

20 – 30 | 25 | 25 | -20 | -2 | -50 |

30 – 40 | 40 | 35 | -10 | -1 | -40 |

40 – 50 | 42 | 45 | 0 | 0 | 0 |

50 – 60 | 33 | 55 | 10 | 1 | 33 |

60 – 70 | 10 | 65 | 20 | 2 | 20 |

Total | \(\sum f_{i}=150\) |
\(\sum (f_{i}\times u_{i})=-37\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 45 – \(\frac{37}{150}\)

= 40 – \(\frac{37}{15}\)

= 45 – 2.466

= 42.534

Thus, the mean is 42.534

Question 6:

In an annual examination, marks (out of 90) obtained by students of Class X in mathematics are given below:

Marks obtained | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |

Number of students | 2 | 4 | 5 | 20 | 9 | 10 |

Find the mean marks.

Solution:

Let us choose a = 52.5, h = 15, then d_{i} = x_{i} – 52.5 and u_{i} = \(\frac{x_{i}-52.5}{15}\)

Using Step-deviation method, the given data is shown as follows:

Marks obtained | Number of students (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 52.5 |
u_{i} = \(\frac{x_{i}-52.5}{15}\) |
f_{i}u_{i} |

0 – 15 | 2 | 7.5 | -45 | -3 | -6 |

15 – 30 | 4 | 22.5 | -30 | -2 | -8 |

30 – 45 | 5 | 37.5 | -15 | -1 | -5 |

45 – 60 | 20 | 52.5 | 0 | 0 | 0 |

60 – 75 | 9 | 67.5 | 15 | 1 | 9 |

75 – 90 | 10 | 82.5 | 30 | 2 | 20 |

Total | \(\sum f_{i}=50\) |
\(\sum (f_{i}\times u_{i})=10\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 52.5 + \(\frac{10}{50}\)

= 52.5 + 3

= 55.5

Thus, the mean is 55.5.

Question 7:

Find the arithmetic mean of the following frequency distribution using step-deviation method:

Age (in years) | 18 – 24 | 24 – 30 | 30 – 36 | 36 – 42 | 42 – 48 | 48 – 54 |

Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |

Solution:

Class | Frequency (f_{i}) |
Mid values (x_{i}) |
u_{i} = \(\frac{x_{i}-A}{h}=\frac{x_{i}-33}{6}\) |
f_{i}u_{i} |

18 – 24 | 6 | 21 | -2 | -12 |

24 – 30 | 8 | 27 | -1 | -8 |

30 – 36 | 12 | 33 = A | 0 | 0 |

36 – 42 | 8 | 39 | 1 | 8 |

42 – 48 | 4 | 45 | 2 | 8 |

48 – 54 | 2 | 51 | 3 | 6 |

Total | \(\sum f_{i}=40\) |
\(\sum (f_{i}\times u_{i})=2\) |

Now, A = 33, h = 6, \(\sum f_{i}=40\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 33 + \(\left \{ 6\times \frac{2}{40} \right \}\)

= 33 + 0.3

= 33.3

Therefore, \(\bar{x}\)

Question 8:

Find the mean of the following data using step-deviation method:

Class | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 | 600 – 620 |

Frequency | 14 | 9 | 5 | 4 | 3 | 5 |

Solution:

Class | Frequency (f_{i}) |
Mid values (x_{i}) |
u_{i} = \(\frac{x_{i}-A}{h}=\frac{x_{i}-550}{20}\) |
f_{i}u_{i} |

500 – 520 | 14 | 510 | -2 | -28 |

520 – 540 | 9 | 530 | -1 | -9 |

540 – 560 | 5 | 550 = A | 0 | 0 |

560 – 580 | 4 | 570 | 1 | 4 |

580 – 600 | 3 | 590 | 2 | 6 |

600 – 620 | 5 | 610 | 3 | 15 |

Total | \(\sum f_{i}=40\) |
\(\sum (f_{i}\times u_{i})=-12\) |

Now, A = 550, h = 20, \(\sum f_{i}=40\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 550 + \(\left \{ 20\times \frac{(-12)}{40} \right \}\)

= 550 – 6

= 544

Therefore, \(\bar{x}\)

Question 9:

Find the mean age from the following frequency distribution:

Age (in years) | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |

Number of persons | 4 | 14 | 22 | 16 | 6 | 5 | 3 |

Solution:

Converting the series into exclusive form, we get:

Class | Frequency (f_{i}) |
Mid values (x_{i}) |
u_{i} = \(\frac{x_{i}-A}{h}=\frac{x_{i}-42}{5}\) |
f_{i}u_{i} |

24.5 – 29.5 | 4 | 27 | -3 | -12 |

29.5 – 34.5 | 14 | 32 | -2 | -28 |

34.5 – 39.5 | 22 | 37 | -1 | -22 |

39.5 – 44.5 | 16 | 42 = A | 0 | 0 |

44.5 – 49.5 | 6 | 47 | 1 | 6 |

49.5 – 54.5 | 5 | 52 | 2 | 10 |

54.5 – 59.5 | 3 | 57 | 3 | 9 |

Total | \(\sum f_{i}=70\) |
\(\sum (f_{i}\times u_{i})=-37\) |

Now, A = 42, h = 5, \(\sum f_{i}=70\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 42 + \(\left \{ 5\times \frac{(-37)}{70} \right \}\)

= 42 – 2.64

= 39.36

Therefore, \(\bar{x}\)

Therefore, Mean age = 39.36 years.

Question 10:

The following table shows the age distribution of patients of malaria in a village during a particular month:

Age (in years) | 5 – 14 | 15 – 24 | 25 – 34 | 35 – 44 | 45 – 54 | 55 – 64 |

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

Find the average age of the patients.

Solution:

Converting the series into exclusive form, we get:

Class | Frequency (f_{i}) |
Mid values (x_{i}) |
u_{i} = \(\frac{x_{i}-A}{h}=\frac{x_{i}-29.5}{10}\) |
f_{i}u_{i} |

4.5 – 14.5 | 6 | 9.5 | -2 | -12 |

14.5 – 24.5 | 11 | 19.5 | -1 | -11 |

24.5 – 34.5 | 21 | 29.5 = A | 0 | 0 |

34.5 – 44.5 | 23 | 39.5 | 1 | 23 |

44.5 – 54.5 | 14 | 49.5 | 2 | 28 |

54.5 – 64.5 | 5 | 59.5 | 3 | 15 |

Total | \(\sum f_{i}=80\) |
\(\sum (f_{i}\times u_{i})=43\) |

Now, A = 29.5, h = 10, \(\sum f_{i}=80\)

Therefore, Mean, \(\bar{x}=A+\left \{ h\times \frac{\sum (f_{i}u_{i})}{\sum f_{i}} \right \}\)

= 29.5 + \(\left \{ 10\times \frac{(43)}{80} \right \}\)

= 29.5 + 5.375

= 34.875

Therefore, \(\bar{x}\)

Therefore, The average age of the patients is 34.87 years.

Question 11:

Weights of 60 eggs were recorded as given below:

Weight (in grams) | 75 – 79 | 80 – 84 | 85 – 89 | 90 – 94 | 95 – 99 | 100 – 104 | 105 – 109 |

Number of eggs | 4 | 9 | 13 | 17 | 12 | 3 | 2 |

Calculate their mean weight to the nearest gram.

Solution:

Let us choose a = 92, h = 5, then d_{i} = x_{i} – 92 and u_{i} = \(\frac{x_{i}-92}{5}\)

Using Step-deviation method, the given data is shown as follows:

Weight (in grams) | Number of eggs (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 92 |
u_{i} = \(\frac{x_{i}-92}{5}\) |
f_{i}u_{i} |

74.5 – 79.5 | 4 | 77 | -15 | -3 | -12 |

79.5 – 84.5 | 9 | 82 | -10 | -2 | -18 |

84.5 – 89.5 | 13 | 87 | -5 | -1 | -13 |

89.5 – 94.5 | 17 | 92 | 0 | 0 | 0 |

94.5 – 99.5 | 12 | 97 | 5 | 1 | 12 |

99.5 – 104.5 | 3 | 102 | 10 | 2 | 6 |

104.5 – 109.5 | 2 | 107 | 15 | 3 | 6 |

Total | \(\sum f_{i}=60\) |
\(\sum (f_{i}\times u_{i})=-19\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 92 + \(\frac{-19}{60}\)

= 92 – 1.58

= 90.42

\(\approx\)

Thus, the mean weight to the nearest gram is 90g.

Question 12:

The following table shows the marks scored by 80 students in an examination:

Marks | Less than 5 | Less than 10 | Less than 15 | Less than 20 | Less than 25 | Less than 30 | Less than 35 | Less than 40 |

Number of students | 3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |

Calculate the mean mark correct to 2 decimal places.

Solution:

Let us choose a = 17.5, h = 5, then d_{i} = x_{i} – 17.5 and u_{i} = \(\frac{x_{i}-17.5}{5}\)

Using Step-deviation method, the given data is shown as follows:

Marks | Number of students (cf) | Frequency (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – 17.5 |
u_{i} = \(\frac{x_{i}-17.5}{5}\) |
f_{i}u_{i} |

0 – 5 | 3 | 3 | 2.5 | -15 | -3 | -9 |

5 – 10 | 10 | 7 | 7.5 | -10 | -2 | -14 |

10 – 15 | 25 | 15 | 12.5 | -5 | -1 | -15 |

15 – 20 | 49 | 24 | 17.5 | 0 | 0 | 0 |

20 – 25 | 65 | 16 | 22.5 | 5 | 1 | 16 |

25 – 30 | 73 | 8 | 27.5 | 10 | 2 | 16 |

30 – 35 | 78 | 5 | 32.5 | 15 | 3 | 15 |

35 – 40 | 80 | 2 | 37.5 | 20 | 4 | 8 |

Total | \(\sum f_{i}=80\) |
\(\sum (f_{i}\times u_{i})=17\) |

The mean of given data is given by

\(\bar{x}=a+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )\times h\)

= 17.5 + \(\frac{17}{80}\)

= 17.5 + 1.06

= 18.56

Thus, the mean mark correct to 2 decimal places is 18.56.