Question

In a physical training camp, students appeared for their per
marks for their physical fitness. They scored below marks out on ou
np. students appeared for their personality test of 80 marks and 20
sical fitness. They scored below marks out of 80 which has been arranged by
wants to find mode but she is unable to find the result. So she asked her teacher for its
solution
Class
0 - 10 10 - 20 | 20 - 30 30-40 40-50 50-60 60 - 70
Frequency 8 T 10 T 10 1 16 | 12 |
borot at the rate of​

Answer 1

Answer:

hshsh hshsh hshshsh isshhs

Answer 2

The mode of the data is 47.89

Step-by-step explanation:

here, The data is not completely given

assuming the data :

class                 frequency

0-10                       8

10-20                     10

20-30                     10

30-40                      1

40-50                      16

50-60                      12

60-70                       3

so , here maximum frequency = 16 thus modal class = 40-50

lower limit (l) = 40

height = 10

f1 = 16

f0= 1

f2 = 12

now ,

mode is

$l+ \frac{f_1-f_0}{2f_1-f_0-f_2}\times h \\\\40 + \frac{16-1}{2\times 16 -1-12}\times 10 \\\\40 + \frac{15}{19}\times 10 \\\\40 + 7.89 \\\\47.89$

hence The mode of the data is 47.89

#Learn more :

Mode+23(mean-mode)mode+23(mean- mode) =

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