Question

If the mean of the following frequency distribution is 188,find the missing frequencies f1 , f2 c.I -0-80,80-160,160-240,240 -320, 320-400 and frequency - 20,25,f1,f2,10 total - 100

Answer 1

Answer:

$f_1=15\:\:and\:\:f_2=30$

Step-by-step explanation:

Given:

Grouped frequency distribution table with 2 missing frequencies.

Mean of the data = 188.

To find: The missing Frequency.

Formula for mean,

$Mean=\frac{\sum_{i=1}^{n}x_if_i}{\sum_{i=1}^{n}f_i}$

Table is attached.

Sum of all Frequency = $55+f_1+f_2$

$55+f_1+f_2=100$

Now,

$188=\frac{7400+200f_1+280f_2}{55+f_1+f_2}$

$10340+188f_1+188f_2=7400+200f_1+280f_2$

$12f_1+92f_2-2940=0$

$3f_1+23f_2-735=0$

Now, let $f_1=x\:\:and\:\:f_2=y$

we have,

x + y = 45 .......................(1)

3x + 23y = 735 .....................(2)

Now, Solving both equation

we get, x = 15 and y = 30

$\implies f_1=15\:\:and\:\:f_2=30$

Therefore, $f_1=15\:\:and\:\:f_2=30$

Answer 2

Answer:

f1=15,f2=30

Step-by-step explanation:

Given : Mean = 188

C.L. fi​ xi​ fi​xi​

0 - 80              20 40              800

80 - 160 25 120 3000

160 - 240 f1​ 200 200 f1​

240 - 320 f2​ 280 280 f2​

320 - 400 10 360 3600

Total ∑fi​=100=55+f1​+f2​  ∑xi​fi​=7400+200f1​+2800f2​

x=∑fi​∑fi​xi​​⇒188=1007400+200f1​+280f2​​

18800=7400+200f1​+280f2​⇒18800−7400=200f1​+280f2​

11400=200f1​+280f2​⇒1140=20f1​+28f2​

285=5f1​+7f2​............(i)

Also f1​+f2​+55=100⇒f1​+f2​=100−55⇒f1​+f2​=45....................(ii)

Solving equation (i) and (ii) f1​=15,f2​=30

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