Math, asked by Anonymous, 10 months ago


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THE MEAN OF THE FOLLOWING DISTRIBUTION OF 100 OBSERVATION IS 148. FIND THE MISSING FREQUENCIES F1 AND F2..

CLASS
0-49
50-99
100-149
150-199
200-249
250-299
300-349

FREQUENCY
10
15
F1
20
15
F2
2

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Answers

Answered by SwaggerGabru
1

Answer:

Step-by-step explanation:

According to question total observation are 100.

So, summation of all frequencies would be 100.

62 + f1 + f2 = 100

f1 + f2 = 100-62

f1 + f2 = 38 --------------eq1

Mean = 148

148 = a + (∑( fi ui)/∑ fi )×h

From attached figure you get that a = 174.5 and h = 50

148 = 174.5 + (-39-f1-2f2 )/( 62+f1+f2 )× 50

(148 -174.5 )÷50 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

-53/100 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

(62+f1+f2 )×(-53) = 100 ×(-39-f1+2f2 )

-3286 -53f1 -53f2 = -3900 -100 f1 +200f2

47 f1 -253 f2 = -614 ------eq2

multiply eq 1 by 47 and subtract both equation

47f1 - 47 f1 + 253 f2 + 47 f2 = 1786 +614

300 f2 = 2400

f2 = 2400/300

f2 = 8

put the value of f2 in equation 1

f1+f2 = 38

f1 = 38 -8

f1 = 30

Answered by Madalasa22
2

Step-by-step explanation:

According to question total observation are 100.

So, summation of all frequencies would be 100.

62 + f1 + f2 = 100

f1 + f2 = 100-62

f1 + f2 = 38 --------------eq1

Mean = 148

148 = a + (∑( fi ui)/∑ fi )×h

From attached figure you get that a = 174.5 and h = 50

148 = 174.5 + (-39-f1-2f2 )/( 62+f1+f2 )× 50

(148 -174.5 )÷50 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

-53/100 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

(62+f1+f2 )×(-53) = 100 ×(-39-f1+2f2 )

-3286 -53f1 -53f2 = -3900 -100 f1 +200f2

47 f1 -253 f2 = -614 ------eq2

multiply eq 1 by 47 and subtract both equation

47f1 - 47 f1 + 253 f2 + 47 f2 = 1786 +614

300 f2 = 2400

f2 = 2400/300

f2 = 8

put the value of f2 in equation 1

f1+f2 = 38

f1 = 38 -8

f1 = 30

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