Question

If the mode of the data is 34.5, find the missing frequency 'f':
Class
15 - 30
BO_46
45 – 60
60 – 75
Frequency
2
7
7​

Answer 1

Step-by-step explanation:

Step-by-step explanation:0+6/2 = 3

Step-by-step explanation:0+6/2 = 3 3*6 = 18

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 9

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 72

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15 15*p = 15 p

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15 15*p = 15 p18+24/2 = 21

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15 15*p = 15 p18+24/2 = 2121*9 = 189

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15 15*p = 15 p18+24/2 = 2121*9 = 18924+30/2 = 27

Step-by-step explanation:0+6/2 = 3 3*6 = 186+12/2 = 99*8 = 7212 +18/ 2 = 15 15*p = 15 p18+24/2 = 2121*9 = 18924+30/2 = 2727*7 =189

Summation of all

Summation of frequency

15.45 = (468 +15p)/30+p

Baaki I think u can solve

If helpful please mark as brainliest

Answer 2

Solution :-

Class Frequency

0 - 15 2

15 - 30 7

30 - 45 f

45 - 60 3

60 - 75 7

we know that,

• Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h

since mode is given as 34.5 . then, modal class is 30 - 45.

so, we have,

• l = lower limit of modal class = 30
• f1 = frequency of modal class = f
• f0 = previous frequency = 7
• f2 = next frequency = 3 .
• h = size of class = 15 .

Putting all values we get,

→ 34.5 = 30 + [(f - 7)/(2f - 7 - 3)] * 15

→ 34.5 - 30 = [(f - 7)/(2f - 10)] * 15

→ 4.5(2f - 10) = 15(f - 7)

→ 9f - 45 = 15f - 105

→ 15f - 9f = 105 - 45

→ 6f = 60

→ f = 10. (Ans.)

Hence, the missing frequency is 10.

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