Well-designed AP 10th Class Maths Guide Chapter 13 Statistics Exercise 13.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Statistics Class 10 Exercise 13.3 Solutions – 10th Class Maths 13.3 Exercise Solutions

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 consumers |

65-85 | 4 |

85 -105 | 5 |

105-125 | 13 |

125-145 | 20 |

145-165 | 14 |

165-185 | 8 |

185-205 | 4 |

Solution:

Median :

Note : Highest frequency class can be treated as the mean class

Median = l + \(\frac{1}{2}\) × h

l = lower limit = 125

N = Total frequency = 68

f = frequency of the median class = 20

h = height of the class = 20

cf = cumulative frequency of the class preceding = 22

= 125 + (34-22) = 125 + 12

Median = 137 units

Therefore, median monthly consumption = 137 units.

Mean :

Note : Highest frequency class can be treated as the mean class

Mean x = a + \(\left(\frac{\sum f_i u_i}{\sum f_i}\right)\) × h

a = class mark of the assumed mean class = 135,

x_{i} = class mark

h = height of the class = 20,

Σf_{i} = Total frequency = 68

Σf_{i}u_{i} = +7

Mean = \(\left(\frac{7}{68}\right)\) = 135 + \(\frac{140}{68}\)= 135 + 2.06 = 137.06 units

Therefore, mean number of units = 137.06 units.

Mode :

Note : Highest frequency class can be treated as the modal class.

\(\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h\)

l = lower limit of modal class = 125

f_{1} = frequency of the modal class = 20

f_{0} = frequency of the class preceding the modal class = 13

f_{2} = frequency of the class succeeding the modal class = 14

h = height of the class = 20

\(\text { Mode }=125+\left[\frac{20-13}{2×20-13-14}\right] \times 20\)

= 125 + \(\left(\frac{7}{40-27}\right)\) × 20

= 125 + \(\frac{140}{13}\)

= 125 + 10.77

= 135.77 units

Mode = 135.77 units

Therefore, the mode of the monthly consumption of number of units = 135.77 units.

The three measures are approximately same in this case.

Question 2.

If the median of the distribution given below is 28.5, find the values 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:

Note : Highest frequency class can be treated as the Median class.

Median = \(l+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right) \times \mathrm{h}\)

l = lower limit = 20,

n = Total frequency = 60

f = frequency of the median class = 20,

h = height of the class = 10

cf = cumulative frequency of the class preceding = 5 + x

Given that N = 45 + x + y = 60

x + y = 60 – 45

x + y = 15 → (1)

= 20 + \(\frac{30-5-x}{2}\) = 28.5

= \(\frac{25-x}{2}\) = 28.5 – 20 = 8.5

25 – x = 8.5 × 2 = 17

-x= 17-25

-x = – 8

x = 8

Put x = 8 in(1) x + y = 15 ⇒ 8 + y = 15 ⇒ y = 15 – 8 = 7 ⇒ y = 7

Therefore, 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 year.

Solution:

Note : Highest frequency class can be treated as the median class.

Median = \(l+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right) \times \mathrm{h}\)

l = lower limit = 35, n = Total frequency = 100,

f = frequency of the median class = 33, h = height of the class = 5

cf = cumulative frequency of the class preceding of the median class = 45

Median = 35 + \(\left(\frac{\frac{100}{2}-45}{33}\right)\) × 5 = 35 + \(\left(\frac{5}{33}\right)\)× 5 = 35 + \(\frac{25}{33}\) = 35 + 0.75 = 35.75

Therefore, median age of the policy holders = 35.75 years.

Question 4.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, 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 the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

Solution:

Mean:

Note : Highest frequency class can be treated as the median class.

Median = \(l+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right) \times \mathrm{h}\)

l = lower limit = \(\frac{144+145}{2}\) = 144.5,

n = Total frequency = 40,

f = frequency of the median class = 12,

h = height of the class = 9

cf = cumulative frequency of the class preceding class of the median class = 17

Median = 144.5 + \(\left(\frac{\frac{40}{2}-17}{12}\right)\) × 9 = 144.5 + \(\left(\frac{20-17}{12}\right)\) × 9

Therefore, median length of the leaf is 147 mm (approximately)

Question 5.

The following table gives the distribution of the life time of 400 neon lamps :

Life time (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:

Note : Highest frequency class can be treated as the median class.

Median = l + \(\left(\frac{\frac{n}{2}-\mathrm{cf}}{\mathrm{f}}\right)\) × h

l = lower ilimit = 3000, N = Total frequency = 400,

f = frequency of the median class = 86, h = height of the class = 500

cf = cumulative frequency of the class preceding class of the median class = 130

Median = 3000 + \(\left(\frac{\frac{400}{2}-130}{86}\right)\) × 500 = 3000 + \(\left(\frac{200-130}{86}\right)\) × 500

= 3000 + (\(\frac{70 \times 500}{86}\)) = 3000 + 406.97

Median = 3407 hour (approx.)

Therefore, median life time of lamp is 3407 hours.

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 sur¬names was obtained as follows:

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:

Median :

Note : High frequency class can be treated as the median class.

Median = l + \(\left(\frac{\frac{n}{2}-\mathrm{cf}}{\mathrm{f}}\right)\) × h

l = lower limit of the median class = 7

n = Total frequency =100

f = frequency of the median class = 40

h = height of the class = 3

cf = cumulative frequency of the class preceding class of the median class = 36

Median = 7 + \(\left(\frac{\frac{100}{2}-36}{40}\right)\) × 3

= 7 + \(\left(\frac{50-36}{40}\right)\) × 3

= 7 + \(\frac{14 \times 3}{40}\) = 7 + 1.05

Median = 8.05 letters

Therefore, median number of letters = 8.05

Mean :

Note : Highest frequency class can be treated as the mean class

Mean \(\overline{\mathrm{E}}\) = a + \(\left(\frac{\sum f_i u_i}{\sum f_i}\right)\) × h

a = class mark of the assumed mean class = 8.5

x_{i} = class mark

h = height of the class = 3

Σf_{i} = Total frequency = 100 Σf_{i}u_{i} = -6

Mean \(\overline{\mathrm{E}}\) = 8.5 + \(\left(\frac{-6}{100}\right)\) × 3

= 8.5 – \(\frac{6 \times 3}{100}\)

= 8.5 – 0.18 = 8.32

Therefore, mean number of letters = 8.32.

Mode :

Note : Highest frequency class can be treated as the modal class.

\(\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h\)

l = lower limit of modal class = 7

f_{1} = frequency of the modal class = 40

f_{o} = frequency of the class preceding the modal class = 30

f_{2} = frequency of the class succeeding the modal class = 16

h = height of the class = 3

Mode = 7 + \(\left(\frac{40-30}{2 \times 40-30-16}\right)\) × 3 = 7 + \(\left(\frac{10 \times 3}{80-46}\right)\) = 7 + \(\frac{30}{34}\) = 7 + 0.88

Mode = 7.88

Therefore, modal number of letter = 7.88.

Question 7.

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

Solution:

Note : Highest frequency class can be treated as the median class.

Median = \(l+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right) \times \mathrm{h}\)

l = lower limit of the median class = 50

n = Total frequency = 30

f = frequency of the median class = 8

h = height of the class = 5

cf = cumulative frequency of the class preceding of the median class = 5

Median = 50 + \(\left(\frac{\frac{30}{2}-5}{8}\right) \times 5\) = 50 + \(\frac{10 \times 5}{8}\) = 50 + 6.25 = 56.25

Median = 56.25 kg.