The mean of 100 observation is 45.It was later found that two observations 19&31 are incorrectly recorded as 91 and 13. Find the correct mean.
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WE KNOW THAT..
MEAN OF THE GIVEN DATA = SIGMA ALL OBSERVATION / NUMBER OF OBSERVATIONS.
LET THE SUM OF ALL OBSERVATION BE X
THEN MEAN =SUM OF ALL OBSERVATION / NUMBER OF OBSERVATIONS.
NOW PUTTING THE VALUES IN THE ABOVE FORMULA.
45 = X / 100
X = 4500
ACCORDING TO QUESTION IT WAS LETTER SEEMED THAT 19 WAS INCORRECTLY WRITTEN AS 91 AND 13 WAS INCORRECTLY WRITTEN AS 31.
WE WILL NOT FIND HOW MUCH THE PREVIOUS OBSERVATION WAS LARGER THAN THE CURRENT OBSERVATION.
SO 91 - 19 = 72
AND 31 - 13 = 18
SHOW THE PREVIOUS DATA WAS LARGER THAN THE CURRENT DATA BY 72 +18 = 90
SUBTRACTING 90 FROM 4500 (SUM OF PREVIOUS OBSERVATIONS)
SUM OF CURRENT OBSERVATION = 4500 - 90 =4410
NOW THE MEAN OF THE CURRENT DATA = 4410 / 100= 44.10
SO THE MEAN OF THE CORRECT DATA IS 44.10.
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MEAN OF THE GIVEN DATA = SIGMA ALL OBSERVATION / NUMBER OF OBSERVATIONS.
LET THE SUM OF ALL OBSERVATION BE X
THEN MEAN =SUM OF ALL OBSERVATION / NUMBER OF OBSERVATIONS.
NOW PUTTING THE VALUES IN THE ABOVE FORMULA.
45 = X / 100
X = 4500
ACCORDING TO QUESTION IT WAS LETTER SEEMED THAT 19 WAS INCORRECTLY WRITTEN AS 91 AND 13 WAS INCORRECTLY WRITTEN AS 31.
WE WILL NOT FIND HOW MUCH THE PREVIOUS OBSERVATION WAS LARGER THAN THE CURRENT OBSERVATION.
SO 91 - 19 = 72
AND 31 - 13 = 18
SHOW THE PREVIOUS DATA WAS LARGER THAN THE CURRENT DATA BY 72 +18 = 90
SUBTRACTING 90 FROM 4500 (SUM OF PREVIOUS OBSERVATIONS)
SUM OF CURRENT OBSERVATION = 4500 - 90 =4410
NOW THE MEAN OF THE CURRENT DATA = 4410 / 100= 44.10
SO THE MEAN OF THE CORRECT DATA IS 44.10.
•••••••••••••••
IF YOU FOUND THIS ANSWER HELPFUL YOU CAN SURELY MARK IT AS THE BRAINLIEST ANSWER...
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