Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.

The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

The class interval having the maximum frequencies is 35 – 45.

f_{1} = 23, l = 35, h = 10, f_{0} = 21, f_{2} = 14

Maximum number of patients admitted in the hospital are of the age 36.8 years. The average age of the patient admitted to the hospital is 35.37 years.

Question 2.

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

Determine the modal lifetimes of the components.

Class interval having the maximum frequency is 60 – 80.

f_{1} = 61, f_{0} = 52, f_{2} = 38, l = 60, h = 20.

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 Rs.) | No. 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:

Step deviation method: \(\bar{x}\) = a + \(\left[\frac{\sum f_i u_i}{\sum f_i}\right]\)h

= 3250 + \(\left[\frac{-235}{200}\right]\) × 500

= 3250 – \(\left[-\frac{1175}{2}\right]\)

= 3250 – 587.5

= 2662.5.

Mean expenditure Rs. 2662.50.

l = 1500, f_{1} = 40, f_{2} = 33, f_{0} = 24, h = 500

Modal monthly expenditure = 1847.83.

Question 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.

Number of students per teacher | No. 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:

l = lower limit of the CI = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3, h = 5

Mode = \(l+\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right] \times \mathrm{h}\)

= \(=30+\left[\frac{10-9}{20-9-3}\right] \times 5\)

= \(30+\left[\frac{1}{8} \times 5\right]=30+\frac{5}{8}\)

= 30 + 0.625 = 30.625.

Most states/UT’s have a student-teacher ratio of 30.6. On an average this ratio is 29.2.

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:

l = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 9, h = 1000

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:

Class interval having the maximum frequency is 40 – 50. f_{1} = 20, f_{0} = 12, f_{2} = 11, l = 40, h = 10.