Exams Joint Entrance Examination Mathematics Statistics Question 17 Question 17 of 208 Medium MCQ
The mean and standard deviation of 100 observations are 40 and 5.1 , respectively. By mistake one observation is taken as 50 instead of 40 . If the correct mean and the correct standard deviation are μ \mu μ and σ \sigma σ
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x 1 , x 2 , … , x 99 , 50 = x 1 + x 2 + … + x 9 + 50 100 = 40 ⇒ x 1 + x 2 + … + x 99 = 4000 − 50 ⇒ x 1 + x 2 + … + x 99 = 3950 = 3950 + 40 100 μ = 399 10 = 39.9 ( ) 2 = ∑ i = 1 99 ( x i ) 2 + 2500 100 − ( 40 ) 2 ∑ i = 1 99 x 2 i = 160101 ( ) 2 = 160101 + 1600 100 − ( 399 10 ) 2 σ = 5 10 ( μ + σ ) = 10 ( 39.9 + 5 ) = 449 \begin{gathered} {x}_{1},{x}_{2},\ldots ,{x}_{99},50 \\ =\frac{{x}_{1}+{x}_{2}+\ldots +{x}_{9}+50}{100}=40 \\ \Rightarrow {x}_{1}+{x}_{2}+\ldots +{x}_{99}=4000-50 \\ \Rightarrow {x}_{1}+{x}_{2}+\ldots +{x}_{99}=3950 \\ =\frac{3950+40}{100} \\ \mu =\frac{399}{10}=39.9 \\ ({)}^{2}=\sum _{i=1}^{99}\frac{{({x}_{i})}^{2}+2500}{100}-(40{)}^{2} \\ \sum _{i=1}^{99}{x}_{2}^{i}=160101 \\ ({)}^{2}=\frac{160101+1600}{100}-{(\frac{399}{10})}^{2} \\ \sigma =5 \\ 10(\mu +\sigma )=10(39.9+5)=449 \end{gathered} x 1 , x 2 , … , x 99 , 50 = 100 x 1 + x 2 + … + x 9 + 50 = 40 ⇒ x 1 + x 2 + … + x 99 = 4000 − 50 ⇒ x 1 + x 2 + … + x 99 = 3950 = 100 3950 + 40 μ = 10 399 = 39.9 ( ) 2 = i = 1 ∑ 99 100 ( x i ) 2 + 2500 − ( 40 ) 2 i = 1 ∑ 99 x 2 i = 160101 ( ) 2 = 100 160101 + 1600 − ( 10 399 ) 2 σ = 5 10 ( μ + σ ) = 10 ( 39.9 + 5 ) = 449 Answer the question to unlock