Question

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect which were recorded as 21,21 and 18. Find the mean and standard deviation if the incorrect observation are omitted​

Answer 1

Answer:

ʜᴇʀᴇ ɪꜱ yᴏᴜʀ ᴀɴꜱᴡᴇʀ..⤵️⤵️

Step-by-step explanation:

qᴜᴇꜱᴛɪᴏɴ :- The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect which were recorded as 21,21 and 18. Find the mean and standard deviation if the incorrect observation are omitted

ᴀɴꜱᴡᴇʀ :- Given, number of observation n=100

Given, number of observation n=100Incorrect mean ( x)=20

Given, number of observation n=100Incorrect mean ( x)=20We know that xˉ = n∑xi

xˉ = n∑xi ⇒20= 1001

xˉ = n∑xi ⇒20= 1001

xˉ = n∑xi ⇒20= 1001

xˉ = n∑xi ⇒20= 1001 i=1

xˉ = n∑xi ⇒20= 1001 i=1∑100xi

xi

xi

xi ⇒ i=l∑100 xi

100 xi

100 xi =20×100=2000

100 xi =20×100=2000So, the incorrect sum of observations =2000.

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean =

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum = 971940 =20

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum = 971940 =20Given, incorrect standard deviation σ=3

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum = 971940 =20Given, incorrect standard deviation σ=3σ= n

100 xi =20×100=2000So, the incorrect sum of observations =2000.Given incorrect observations are 21,21,18 and these has to be omitted.So, correct sum of observations =2000−21−21−18=2000−60=1940Correct mean = 100−3Correctsum = 971940 =20Given, incorrect standard deviation σ=3σ= n∑i=1n xi²

−(n∑i=1ᴍɴxi)2

22 =n∑

22 =n∑ i=1

22 =n∑ i=1nxɪ2 −(xˉ) 2

2 −(xˉ) 2

2 −(xˉ) 2 ⇒32

2 −(xˉ) 2 ⇒32 = 100∑x ɪ²

−(20)2

−(20)2

−(20)2 ⇒Incorrect∑xi2

−(20)2 ⇒Incorrect∑xi2

−(20)2 ⇒Incorrect∑xi2 =100(9+400)=40900

−(20)2 ⇒Incorrect∑xi2 =100(9+400)=40900Correct

−(20)2 ⇒Incorrect∑xi2 =100(9+400)=40900Correct i=∑ɴ xi2 =Incorrect

xi2 =Incorrect i=l

xi2 =Incorrect i=l∑n xi2

xi2 =Incorrect i=l∑n xi2 −(21)²

−(21)² −(18)²

=40900−441−441−324

=40900−441−441−324=39694

=40900−441−441−324=39694∴ Correct standard deviation = n correct∑xi2 −(Correctmean)2

correct∑xi2 −(Correctmean)2

correct∑xi2 −(Correctmean)2

correct∑xi2 −(Correctmean)2

correct∑xi2 −(Correctmean)2 = 9739694 −(20)²

=409.216−400

=409.216−400

=409.216−400

=409.216−400 = 9.216

=409.216−400 = 9.216

=409.216−400 = 9.216

=409.216−400 = 9.216 =3.036

ʜᴏᴩᴇ ɪᴛꜱ ʜᴇʟᴩ..⤴️⤴️

Answer 2

Answer:

Incorrect mean is 20 . There was 100 observations .

So sum of all observations = 20 × 100 = 2000

$2000 - 21 - 21 - 18 = 1940$

Correct Mean =

$\frac{1940}{100 - 3} = \frac{1940}{97} = 20$

Now , we will find the correct Standard Deviation

We know ,

$σ = \sqrt { \frac{1}{n}Σ(xi) {}^{2} - (x) {}^{2} }$

When incorrect observations were present , the standard deviation was 9

$3 = \sqrt{ \frac{1}{100} Σ(xi) {}^{2} - (20) {}^{2} }$

$9 = \frac{1}{100} Σ(xi) {}^{2} -400$

$40900 = Σ(xi) {}^{2}$

This is the sum when observations were incorrect .

Correct sum when observations are omitted $40900 - 21 {}^{2} - 21 {}^{2} - 18 {}^{2} = 39694$

Correct $Σ(xi) {}^{2} = 39694$

$σ = \sqrt{ \frac{1}{97} (39694) - 400} = \sqrt{409.16 - 400} = \sqrt{9.16} = 3.03$

So , the correct deviation is 3.03 .