The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. Find the correct mean.

Question

The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. Find the correct mean. (A) 38.9 (B) 37.9 (C) 39.9 (D) 36.9

Tardigrade Admin · Accepted Answer

Given that number of observations (n) = 100 Incorrect mean ( barx) = 40 We know that barx =(1/n)∑ limitsi=1nxi i.e. 40= (1/100)∑ limitsi=1100 xi or ∑ limitsi=1100xi = 4000 i.e. Incorrect sum of observations = 4000 Thus the correct sum of observations = 4000 - 50 + 40 = 3990 Hence Correct Mean = (3990/100) = 39.9

Q. The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ , respectively by a student who took by mistake $50$ instead of $40$ for one observation. Find the correct mean.

$38.9$

$37.9$

$39.9$

$36.9$

Q. The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. Find the correct mean.

38.9

37.9

39.9

36.9

Given that number of observations (n)=100 Incorrect mean (xˉ)=40 We know that xˉ=n1​i=1∑n​xi​ i.e. 40=1001​i=1∑100​xi​ or i=1∑100​xi​=4000 i.e. Incorrect sum of observations =4000 Thus the correct sum of observations =4000−50+40=3990 Hence Correct Mean =1003990​ =39.9

Q. The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ , respectively by a student who took by mistake $50$ instead of $40$ for one observation. Find the correct mean.

$38.9$

$37.9$

$39.9$

$36.9$