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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Statistics

NTA Abhyas 2020 · Mathematics · Statistics

The mean of five observations is 4 and their variance is 2.8 . If...

The mean of five observations is 44 and their variance is 2.82.8. If three of these observations are 2,22, 2 and 55, then the other two are

Options

  1. A22 and 99
  2. B33 and 88
  3. C44 and 77
  4. D55 and 66

Correct answer

D. 55 and 66

Step-by-step solution

x¯=4,N=5\overset{¯}{x} = 4, N = 5
and Σ(x−x¯)2N=2.8\dfrac{Σ{\left(x - \overset{¯}{x}\right)}^{2}}{N} = 2.8
⇒Σ(x−x¯)2=(2.8)5⇒ Σ{\left(x - \overset{¯}{x}\right)}^{2} = \left(2.8\right)5
∴Σ(x−x¯)2=14∴ Σ{\left(x - \overset{¯}{x}\right)}^{2} = 14
∴(2−4)2+(2−4)2+(5−4)2+(α−4)2+(β−4)2=14∴ {\left(2 - 4\right)}^{2} + {\left(2 - 4\right)}^{2} + {\left(5 - 4\right)}^{2} + {\left(\alpha - 4\right)}^{2} + {\left(\beta - 4\right)}^{2} = 14
Where α,β\alpha, \beta are the other two observations.
∴4+4+1+(α−4)2+(β−4)2=14∴ 4 + 4 + 1 + {\left(\alpha - 4\right)}^{2} + {\left(\beta - 4\right)}^{2} = 14
∴(α−4)2+(β−4)2=5∴ {\left(\alpha - 4\right)}^{2} + {\left(\beta - 4\right)}^{2} = 5
Also, 2+2+5+α+β5=4\dfrac{2 + 2 + 5 + \alpha + \beta}{5} = 4
∴α+β=20−9=11∴ \alpha + \beta = 20 - 9 = 11
Clearly 5,65, 6 only satisfy the above equations in α,β.\alpha, \beta .
Hence, the required numbers are 5,6.5,6 .
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