Question

the median of the following data is 49.2. Find the values of f1 and f2 if summation fi is 90:
ne median of the follo
Marks
20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 - 80 80 - 90
frequency
f1 15 25 20 f2 8 10
​

Answer 1

Answer:

the median of the following data is 49.2 . The values of f1 and f2 are 7 and 5 respectively , if £fi =90  marks – 20-30, 30-40,40-50, 50-60, 60-70, 70-80,80-90  Frequency – f1 , 15,25, 20, f2, 8, 10  

 Step-by-step explanation: Given that median=49.2 From the table (shown below), we get f1+15+25+20+f2+8+10=90. That is, f1+f2=12. Formula for   where, L is the lower limit of the median class f is the frequency of the median class cf is the cumulative frequency of the class preceding to the median class i is the width of the median class N is the sum of all frequencies The median class being the class which contains the (N/2)th observation. Here the median class is 40-50, since the median is 49.2 so that  L=40, N=90, cf=f1+15, f=25 and i=10 putting these values in the equation of median we get,

Answer 2

f₁ = 7 and f₂ = 5

Step-by-step explanation:

N = ∑fi = 90

(15 + 25 + 20 + 8 + 10) + f₁ + f₂ = 90

78 + f₁ + f₂ = 90

f₁ + f₂ = 90 - 78

f₁ + f₂ = 12

Median = 49.2

∴ Median class = value of $(\frac{n}{2})th$

                         = $(\frac{90}{2} )th$

                          = 45th observation

We find that 45th observation lies in the class 40-50

Therefore, median class = 40-50

Median = $M=L+(\frac{\frac{n}{2}-cf }{f} )\times c$

L=lower boundary point of median class = 40

n=Total frequency = 90

cf=Cumulative frequency of the class preceding the median class = f₁+15

f=Frequency of the median class = 25

c=class length of median class = 10

$49.2=40+(\frac{45-f_1-15}{25} )\times 10$

$49.2-40=\frac{30-f_1}{25}\times 10$

$9.2\times \frac{25}{10}=30-f_1$

23 = 30 - f₁

f₁ = 30 - 23

f₁ = 7

As per equation putting the value of f₁

f₁ + f₂ = 12

7 + f₂ = 12

f₂ = 12 - 7

f₂ = 5

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