The mean and variance of seven observations are 8 \; and \; 16, respectively. If 5 of the observations are 2,4,10,12,14, then the product of the remaining two observations is :
 

  • Option 1)

    49

  • Option 2)

    48

  • Option 3)

    45

     

  • Option 4)

    40

 

Answers (1)

Mean \bar{a}=\frac{\sum x_ i}{n}

Variance =\sigma ^{2}=\frac{\sum x_i^{2}}{n}-(\bar{x})^{2}

Given \bar{x}=8

\bar{x}=\frac{2+4+10+12+14+a+b}{7}=8

=\frac{16+26+a+b}{7}=8

=42+a+b=56

a+b=14\; \; \; (1)

\sigma ^{2}=\frac{4+16+100+144+196+a^{2}+b^{2}}{7}-64=16

=460-448+a^{2}+b^{2}=112

=12+a^{2}+b^{2}=112

a^{2}+b^{2}=100

(a+b)^{2}=a^{2}+b^{2}+2ab

\\a+b=14 \; \; \; \; \; from (1)\\\; \; \; \\a^{2}+b^{2}=100\; \; \; \; \; \; \; \; from (11)

\\196=100+2ab\\\; \; \; \\96=2ab\\\; \; \; \\ab=48


Option 1)

49

Option 2)

48

Option 3)

45

 

Option 4)

40

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