Well-designed AP 10th Class Maths Guide Chapter 13 Statistics Exercise 13.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Statistics Class 10 Exercise 13.2 Solutions – 10th Class Maths 13.2 Exercise Solutions

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 |

No. 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:

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 = 35

f_{1} = frequency of the modal class = 23

f_{0} = frequency of the class preceding the modal class = 21

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

h = height of the class = 10

\(\text { Mode }=35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times 10\)

= 35 + \(\frac{2}{46-35}\) × 10 = 35 + \(\frac{20}{11}\) = 35 + 1.81

∴ Mode = 36.81

Mean:

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

Mean=\(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

a = class mark of the assumed mean class = 40

x_{i} = class mark

h = height of the class = 10

Σf_{i} = Total frequency = 80

Σf_{i}u_{i} = -37

Mean \(\overline{\mathrm{x}}\) = 40 + \(\frac{-37}{80}\) × 10 = 40 – \(\frac{37}{8}\)

= 40 – 4.625

Mean = 35.375

∴ Mode is more than mean.

Question 2.

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

Determaine the model of lifetimes of components.

Solution:

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 = 60

f_{1} = frequency of the modal class = 61

f_{0} = frequency of the class preceding the modal class = 52

f_{2} = frequency of the class succeeding the modal class = 38

h = height of the class = 20

\(\text { Mode }=60+\left[\frac{61-52}{2 ×62-52-38}\right] \) × 20

= 60 + (\(\frac{90×20}{32}\))

= 60 + 5.625

= 65.625

∴ Modal lifetime of the components = 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 (in ₹) | 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:

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 = 1500

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

f_{0} = frequency of the class preceding the modal class = 24

f_{2} = frequency of the class succeeding the modal class = 33

\(\text { Mode }=1500+\left[\frac{40-24}{2×40-24-33}\right]\) × 500

= 1500 + \(\frac{16×500}{23}\)

= 1500 + 347.82 = 1847.82 /-

∴ Modal monthly expenditure = ₹ 1847.82 /-

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

Mean \(\overline{\mathbf{x}}=a+\frac{\sum f_i u_i}{\sum f_i} \times h\)

a = class mark of the assumed mean class = 1750

x_{i} = class mark

h = height of the class = 500

Σf_{i} = Total frequency = 200

Σf_{i}u_{i} = 365

= 1750 + 912.50

Mean = ₹ 2662.50

Therefore, mean monthly expenditure = ₹ 2662.50

Question 4.

The following distribution gives the state-wise teacher-student ratio in higher sec¬ondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:

Mean :

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

Mean \(\overline{\mathbf{x}}=a+\frac{\sum f_i u_i}{\sum f_i} \times h\)

a = class mark of the assumed mean class = 32.5

x_{i} = class mark

h = height of the class = 5

Σf_{i} = Total frequency = 35

Σf_{i}u_{i} = -23

Therefore, mean number of students per each teacher = 30.625 (or) 31 (Approximately)

Mode:

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

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

l = lower limit of the modal class = 30

f_{1} = frequency of the modal class = 10

f_{0} = frequency of the class preceding the modal class = 9

f_{2} = frequency of the class succeeding the modal class = 3

h = height of the class = 5

\(\text { Mode }=30+\left[\frac{10-9}{2×10-9-3}\right]\) × 5

= 30 + \(\frac{1}{20-12}\) × 5

= 30 + \(\frac{1}{8}\) = 30 + 0.625

Mode = 30.625

Therefore, modal number of students per each teacher = 30.625 (or) 31 (Approximately)

Question 5.

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

Runs scored | Number of batsmen |

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:

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

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

l = lower limit of the modal class = 4000

f_{1} = frequency of the modal class = 18

f_{0} = frequency of the class preceding the modal class = 4

f_{2} = frequency of the class succeeding the modal class = 9

h = height of the class = 1000

\(\text { Mode }=4000+\left[\frac{18-4}{2×18-4-9}\right]\) × 1000

= 4000 + \(\frac{14}{36-13}\) × 1000 = 4000 + (\(\frac{14}{23}\)) × 1000

= 4000 + 608.69 = 4608.69

Mode = 4608.69 (or) 4609 runs (approx.)

Therfore, mode = 4609 runs approximately.

Question 6.

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:

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

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

l = lower limit of the modal class = 40

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

f_{0} = frequency of the class preceding the modal class = 12

f_{2} = frequency of the class succeeding the modal class = 11

h = height of the class = 10

\(\text { Mode }=40+\left[\frac{20-12}{2×20-12-11}\right]\) × 10

= 40 + \(\frac{8}{40-23}\) × 10 = 40 + \(\frac{80}{17}\) = 40 + 4.706 = 44.706 cars

Therefore, mode of the data = 44.706 cars.