Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(c) can lead to a stronger grasp of mathematical concepts.

## S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(c)

Question 1.

Find the mode of the following data :

(i) 3, 4, 7, 11, 4, 3, 4, 5, 6, 4, 1, 2, 4, 4

(ii) Size of shoes : 4, 4.5, 5, 4.5, 5.5, 5, 6, 4.5, 4, 4.5

(iii) Wages (₹) : 100, 120, 100, 120, 130, 120, 120, 130, 120, 100

(iv) Runs in an innings : 18, 32, 0, 40, 60, 69, 33, 69, 35, 11, 20

Solution:

(i) Arranging the given data in ascending order; we have

1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 11

Here 4 repeated maximum no. of times.

∴ required mode = 4

(ii) Arranging the given data in ascending order : 4, 4, 4.5, 4.5, 4.5, 4.5, 5, 5, 5.5, 6

Here 4.5 repeated maximum no. of times.

∴ required mode = 4.5

(iii) Arranging the given data in ascending order :

100, 100, 100, 120, 120, 120, 120, 120, 130, 130

Here 120 repeated maximum no. of times i.e. 5 times. Thus required mode = 120

(iv) Arranging the given data in ascending order: 0, 11, 18, 20, 32, 33, 35, 40, 60, 69, 69

Here 69 repeated maximum no. of times

∴ required mode = 69

Question 2.

Find the mode from the following data in question 2 to 7.

Marks | 10 | 12 | 15 | 22 | 25 | 35 | 45 | 50 | 60 |

Number of students | 4 | 6 | 10 | 14 | 20 | 19 | 10 | 6 | 3 |

Solution:

Here maximum frequency be 20 and corresponding observation be 25. Thus required modal marks be 25.

Question 3.

Size | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |

Frequency | 3 | 5 | 1 | 7 | 8 | 6 | 4 | 2 | 3 |

Solution:

The table of values is given as under:

∴ Mean = \(\frac{\Sigma f x}{\Sigma f}\) = \(\frac{652}{41}\)

Here, \(\frac{N+1}{2}\) = \(\frac{41+1}{2}\) = 21

Thus M_{d} = 16

∴ Mode = 3 median – 2 mean = 3 × 16 – 2 × \(\frac{652}{41}\) ≃ 16.19

Question 4.

Class interval | 0-10 | 10-20 | 20-30 | 30-40 |

Frequency | 10 | 45 | 12 | 3 |

Solution:

From given data, maximum frequency be 45 and thus 10 – 20 be the modal class.

Here l = 10; f_{m} = 45 ; f_{m-1} = 10; f_{m+1} = 12 ; i = 10

Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 10 + \(\frac{45-10}{90-10-12} \times 10\) = 10 + \(\frac{350}{68}\) = 10 + 5.1470 = 15.147

Question 5.

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Number of students | 8 | 16 | 26 | 60 | 38 |

Solution:

Here maximum frequency be 60 and modal class be 30 – 40.

Here, l = 30; f_{m} = 60; f_{m-1} = 28 ; f_{m+1} = 38 ; i = 10

∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 30 + \(\frac{60-28}{120-28-38} \times 10\) =30 + \(\frac{320}{56}\) = 35.71

Question 6.

Marks | Frequency | Marks | Frequency |

1-5 | 7 | 26-30 | 18 |

6-10 | 10 | 31-35 | 10 |

11-15 | 16 | 36-40 | 5 |

16-20 | 32 | 41-45 | 1 |

21-25 | 24 |

Solution:

Given class-intervals are discontinuous we make it continuous with the help of adjustment factor.

Here adjustment factor = \(\frac{6-5}{2}\) = 0.5

We subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Here modal class be 15.5 – 20.5

∴ l = 15.5; f_{m} = 32 ; f_{m-1} = 16; f_{m+1} = 24; i = 5

Thus,

Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\)

= 15.5 + \(\frac{32-16}{64-16-24} \times 5\)

= 15.5 + \(\frac{80}{24}\) ≃ 18.84

Question 7.

Mid-value | 60 | 64 | 68 | 72 | 76 |

Frequency | 25 | 8 | 10 | 6 | 4 |

Solution:

The table of values is given as under:

Mid-value | Class Intervals | Frequency(f) |

60 | 58-62 | 25 |

64 | 62-66 | 8 |

68 | 66-70 | 10 |

72 | 70-74 | 6 |

76 | 74-78 | 4 |

Σf = N = 53 |

Here Modal class be 58 – 62.

Here

l = 58; f_{m} = 25 ; f_{m+1} = 8 ; i = 4 ; f_{m-1} = 0

Thus Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 58 + \(\frac{25-0}{50-8}\) × 4 = 60.38

Question 8.

(i) The mode of the following frequency distribution is 48.6 Find the unknown frequency.

Class | 20-25 | 25-40 | 40-55 | 55-70 | 70-85 |

Frequency | 6 | 20 | 44 | ? | 3 |

(ii) If the frequency of the class 70-85 is 13 instead of 3, then what difference will it make?

Solution:

Let the missing frequency be f.

Here modal class be 40 – 55.

∴ l = 40; f_{m} = 44 ; f_{m-1} = 20 ; f_{m+1} = f; i = 15

Thus,

Mode = \( l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m-1}} \times i\)

48.6 = 40 + \(\frac{44-20}{88-20-f} \times 15\)

⇒ 8.6 = \(\frac{360}{68-f}\)

⇒ 584.8 – 8.6f = 360

⇒ 8.6(68 – f) = 360

⇒ 584.8 – 8.6f = 360

⇒ 8.6 f = 224.8

⇒ f = 26.14

If the frequency of class 70 – 85 is 13 instead of 3.

Then value of mode is unaffected as value of mode affected if there are changes in the values of f_{m}, f_{m-1} and f_{m+1}.

Question 9.

Find the mean, median and mode of the following :

(i) The data 17, 32, 35, 33, 15, 21, 41, 32, 11, 18, 20, 22, 11, 15, 35, 23, 38, 12.

(ii)

Size of item | 11 | 12 | 13 | 14 | 15 | 16 | 17 |

Frequency | 8 | 13 | 25 | 37 | 23 | 15 | 9 |

Solution:

(i) Required Mean = \(\frac{17+32+35+33+15+21+41+32+11+18+20+22+11+15+35+23+38+12}{18}\)

= \(\frac{431}{18}\) = 23.94

Arranging the given data in ascending order ; we have

11, 11, 12, 15, 15, 17, 18, 20, 21, 22, 23, 32, 32, 33, 35, 35, 38, 41

Here no. of observations = n = 18(even)

The observations 11, 15, 32 and 35 repeated equal no. of times i.e. 2 times. Thus, mode is illdefined.

(ii) The table of values is given as under:

∴ Mean = \(\frac{\Sigma f x}{\Sigma f}\) = \(\frac{1825}{130}\) ≃ 14.04

Here, \(\frac{N+1}{2}\) = \(\frac{130+1}{2}\) = 65.5 ∴ Median = 14

Further maximum frequency be 37 and corresponding value of x i.e. 14 gives the required value of mode.

Question 10.

The mode of the following distribution is 240. Find out the missing frequency :

Size | Frequency | Size | Frequency |

0-100 | 140 | 300-400 | ? |

100-200 | 230 | 400-500 | 150 |

200-300 | 270 | 500-600 | 140 |

Solution:

Let the missing frequency be f.

Clearly the maximum frequency be 270 and modal class be 200 – 300.

Here l = 200; f_{m} = 270 ; f_{m-1} = 230 ; f_{m+1} = f ; i = 100

Thus, Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) ⇒ 240 = 200 + \(\frac{(270-230)}{540-230-f} \times 100\)

⇒ 40 = \(\frac{4000}{310-f}\) ⇒ 40(310 – f) = 4000 ⇒ 12400 – 40f = 4000 ⇒ 40f = 8400 ⇒f = 210

Question 11.

Find the median, mode, third quartile, 5th decile and 65th percentile from the following :

Value of the item | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

Frequency | 25 | 30 | 38 | 45 | 46 | 58 | 32 | 34 | 25 | 22 |

Solution:

The table of values is given as under:

Here \(\frac{N+1}{2}\) = \(\frac{355+1}{2}\) = 178 ∴ M_{d} = 5

Here maximum Frequency be 58 and corresponding value of x i.e. 6 be the required value of mode.

Q_{3} : Here \(\frac{3(\mathrm{~N}+1)}{4}\) = \(\frac{3 \times 356}{4} \) = 267 and corresponding c.f be 274 and the corresponding value of x gives the required value of Q_{3}. Thus Q_{3} = 7

D_{5} : Here, \(\frac{5(\mathrm{~N}+1)}{10}\) = \(\frac{5 \times 356}{10}\) = 178

∴ D_{6} = size of 178th item = 5

P_{65 }; Here, \(\frac{65(\mathrm{~N}+1)}{100}\) = \(\frac{65 \times 356}{100}\) = 231.4

∴ P_{65} = size of 231.4th item = 6

Question 12.

Find median, mean and modal marks and also determine the limits between which 80% of the students have secured marks.

Marks | No. of students | Marks | No. of students |

0-10 | 15 | 50-60 | 80 |

10-20 | 25 | 60-70 | 70 |

20-30 | 52 | 70-80 | 10 |

30-40 | 56 | 80-90 | 5 |

40-50 | 78 | 90-100 | 2 |

Solution:

The table of values is given as under:

Then by Step-deviation method, we have

Mean \(\bar{x}\) = A + \(\frac{\Sigma f d^{\prime}}{\mathrm{N}}\) × i = 45 – \(\frac { 15 }{ 393 }\) × 10 = 44.62

Here \(\frac { N }{ 2 }\) = \(\frac { 393 }{ 2 }\) = 196.5 which lies in 40 – 50.

Thus median class be 40 – 50.

Here

l = 40; f = 78; c = 148 ; i = 10

∴ M_{d} = \(l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times i\) = 40 + \(\frac{196.5-148}{78} \times 10\) = 40 + 6.2195 = 46.22

Clearly modal class be 50 – 60.

Here l = 50 ; f_{m} = 80 ; f_{m-1} = 78 ; f_{m+1} = 70 ; i = 10

∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 50 + \(\frac{80-78}{160-78-70}\) × 10 = 50 + \(\frac { 20 }{ 12 }\) = 51.67

P_{10}: Here \(\frac{10 \times N}{100}\) = \(\frac{10 \times 393}{100}\) = 39.3 which lies in 10 – 20

Here l = 10; c = 15; f = 25; i = 10

\(\mathrm{P}_{90}=l+\frac{\frac{90 \mathrm{~N}}{100}-\mathrm{C}}{f} \times i\) = \(60+\frac{353.7-306}{70} \times 10\) = 66.81

The required limits between which 80% students have secured marks be 19.72 and 66.81.