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JEE Main 202126 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2021 · 26 August · Morning Shift · Mathematics · Statistics

The mean and standard deviation of 20 observations were calculated...

The mean and standard deviation of 2020 observations were calculated as 1010 and 2.52 . 5 respectively. It was found that by mistake one data value was taken as 2525 instead of 35.35 . If α\alpha and β\sqrt{\beta} are the mean and standard deviation respectively for correct data, then (α,β)( \alpha, \beta ) is:

Options

  1. A(10.5,26)( 10 . 5, 26 )
  2. B(10.5,25)( 10 . 5, 25 )
  3. C(11,25)( 11, 25 )
  4. D(11,26)( 11, 26 )

Correct answer

A. (10.5,26)( 10 . 5, 26 )

Step-by-step solution

Mean =x1+x2+x3+……+x19+2520=10= \dfrac{{x}_{1} + {x}_{2} + {x}_{3} + \ldots \ldots + {x}_{19} + 25}{20} = 10⇒x1+x2+x3+………+x19+25=200⇒ {x}_{1} + {x}_{2} + {x}_{3} + \ldots \ldots \ldots + {x}_{19} + 25 = 200

⇒x1+x2+x3+………+x19=175⇒ {x}_{1} + {x}_{2} + {x}_{3} + \ldots \ldots \ldots + {x}_{19} = 175

New mean =x1+x2+x3+……+x19+3520=175+3520=10.5= \dfrac{{x}_{1} + {x}_{2} + {x}_{3} + \ldots \ldots + {x}_{19} + 35}{20} = \dfrac{175 + 35}{20} = 10 . 5

S.D. =x12+x22+x32+……+x192+(25)220−(10)2= \sqrt{\dfrac{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + \ldots \ldots + {x}_{19}^{2} + ( 25{)}^{2}}{20} - ( 10{)}^{2}}

⇒2.5=x12+x22+x32+……+x192+(25)220−(10)2⇒ 2 . 5 = \sqrt{\dfrac{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + \ldots \ldots + {x}_{19}^{2} + ( 25{)}^{2}}{20} - ( 10{)}^{2}}

⇒x12+x22+x32+……+x192=1500⇒ {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + \ldots \ldots + {x}_{19}^{2} = 1500

New S.D.=x12+x22+x32+……+x192+(35)220−(10.5)2S . D . = \sqrt{\dfrac{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + \ldots \ldots + {x}_{19}^{2} + ( 35{)}^{2}}{20} - ( 10 . 5{)}^{2}}

=1500+(35)220−(110.25)= \sqrt{\dfrac{1500 + ( 35{)}^{2}}{20} - ( 110 . 25 )}

=26= \sqrt{26}

(α,β)=(10.5,26)( \alpha, \beta ) = ( 10 . 5, 26 )

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