Question

Arithmetic mean and s.d. Of 12 items are 22 and 3 respectively later on it was observed that the item 32 was wrongly taken as 23 computer correct mean, S.D. and C.V.

Answer 1

Answer:

tapon mo nalang module mo

Answer 2

Answer:

Correct mean = 22.75

Standard deviation = 4.0850

Coefficient of variance = 17.956

Step-by-step explanation:

Given data

Arithmetic mean of 12 items = 22

standard deviation of the data = 3

the incorrect observation  = 23

the correct observation = 32  

here we need to find correct mean, S.D and coefficient of variance of the given data  

from given data Arithmetic mean x =  ∑x/ n = 22

          here "n" is number of items  

                   "∑x" is sum of 12 items  

⇒    ∑x/ 12 = 22    

⇒    ∑x = 12 × 22 = 264

here 32 is written as 23

corrected   ∑x = 264 - 23 + 32 = 273  

∴ corrected mean  x = 273/ 12 = 22.75

given standard deviation S.D = √(∑x²/ n - mean²) = 3

         here ∑x² = sum of squares of observations

⇒   √(∑x²/ n - mean²) = 3  

⇒    (√∑x²/12 - 22²)² = 3²  [ squaring on both sides]  

⇒     ∑x²/ 12 - 22² = 9  

⇒     ∑x²/12 = 9 + 484

⇒      ∑x² = 493×12 = 5916            

in ∑x²,  32² is written as 23²  

corrected  ∑x² = 5916 - 23² +  32²

                        = 5916 - 529 + 1024

                 ∑x² = 6411    

∴ corrected standard deviation S.D = √(∑x²/ n - mean²)  

                                                           = √[6411/12 - (22.75)² ]

                                                           = √(534.25 - 517.5625)                

                                                           = √16.6875 = 4.0850 (approx)

∴ coefficient of variance C.V = $\frac{S.D}{mean } (100)$

                                                = $\frac{4.0850}{22.75} (100) = 17.956$