Question

0. In annual day of a school, age-wise participation of students is shown in the ...
09-Jan-2020 · In annual day of a school, age-wise participation of students is shown in the following frequency distribution:Age of student (in years) 5-7 | 7-9 9-11 11 – 13 13-15 15–17 | 17-19Number of students 20 i 18 22 25 20 15 10Draw a 'less than type ogive for the above data and from it find the median age

Answer 1

11.4 is median

Step-by-step explanation:

Age                  f                    cf

5 - 7                 20                20

7 - 9                18                  38

9 - 11                22                 60

11 - 13               25                 85

13 - 15              20                105

15 - 17              15                  120

17 - 19               10                 130

Median =  

11 +  2 * ( 65 - 60)/25

= 11 + 0.4

= 11.4

Less than ogive

The points with the upper limits of the class are plotted on x axis  and the corresponding less than cumulative frequencies on y axis.

The points are joined by free hand smooth curve to give less than cumulative frequency curve or the less than Ogive

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Answer 2

The median is 11.4

Step-by-step explanation:

In annual day of a school, age-wise participation of students is shown given below

:Age of student             Number of students (f)         cf

less than 7                            20                                   20

less than 9                            18                                    38

less than 11                             22                                   60

less than 13                            25                                   85

less than 15                             20                                  105

less than 17                             15                                    120

less than 19                            10                                     130

then

here median class is 11-13 , h = 2 , n/2 = 65 , cf = 60 f= 25

median = $l+ \frac{\frac{n}{2}-cf }{f}\times h$

median = $11 + \frac{65-60}{25}\times 2$

median = 11.4

hence , The median is 11.4

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