# Consider the following frequency distribution:

Class 0-5 6-11 12-17 18-23 24-29

Frequency 13 10 15 8 11

The upper limit of the median class is

a. 17

b. 17.5

c. 18

d. 18.5

**Solution:**

As the classes are not continuous, we have to make the data continuous by subtracting 0.5 from lower limit and adding 0.5 to the upper limit of each class

Class | Frequency | Cumulative Frequency |

0-5.5 | 13 | 13 |

6.5-11.5 | 10 | 23 |

11.5-17.5 | 15 | 38 |

17.5-23.5 | 8 | 46 |

23.5-29.5 | 11 | 57 |

From the question,

N/2 = 57/2 = 28.5

We know that 28.5 lies between the interval 11.5-17.5

So the upper limit is 17.5

Therefore, the upper limit of the median class is 17.5.

**✦ Try This: **Consider the following frequency distribution:

Class | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 |

Frequency | 12 | 10 | 5 | 9 | 13 |

The upper limit of the median class is

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 14

**NCERT Exemplar Class 10 Maths Exercise 13.1**** Problem 7**

## Consider the following frequency distribution: Class 0-5 6-11 12-17 18-23 24-29 Frequency 13 10 15 8 11. The upper limit of the median class is a. 17, b. 17.5, c. 18, d. 18.5

**Summary:**

Considering the frequency distribution, the upper limit of the median class is 17.5

**☛ Related Questions:**

- For the following distribution: Marks Number of students Below 10 3 Below 20 12 Below 30 27 Below 40 . . . .
- Consider the data: Class 65-85 85-105 105-125 125-145 145-165 165-185 185-205 Frequency 4 5 13 20 14 . . . .
- The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below: Class 1 . . . .

Math worksheets and

visual curriculum

visual curriculum