Let 9 = x 1 < x 2 < … < x 7 be in an A.P. with common difference d . If the standard deviation of x 1 , x 2 … , x 7 is 4 and the mean is x ¯ , then x ¯ + x 6 is equal to :

Question

Let 9 = x 1 &#60; x 2 &#60; &#8230; &#60; x 7 be in an A.P. with common difference d . If the standard deviation of x 1 , x 2 &#8230; , x 7 is 4 and the mean is x &#175; , then x &#175; + x 6 is equal to :

Quantrex Academy · Accepted Answer

Correct answer: B. 34

Given, 9 = x 1 &#60; x 2 &#60; &#8230; &#60; x 7 be in an A.P. with common difference d And the standard deviation of x 1 , x 2 &#8230; , x 7 is 4 and the mean is x &#175; , Now solving, 9 = x 1 &#60; x 2 &#60; &#8230; &#8230; &#60; x 7 which is an A.P, we get, 9 , &#160; 9 + d , &#160; 9 + 2 d , &#8230; &#8230; &#8230; . 9 + 6 d Now subtracting 9 from the series we get, 0 , &#160; d , &#160; 2 d , . . . . . . . 6 d So, mean will be x &#175; new = 21 d 7 = 3 d Now using the formula of variance we get, &#963; 2 = &#

Let 9 = x 1 < x 2 < … < x 7 be in an A.P. with common difference d . If the standard deviation of x 1 , x 2 … , x 7 is 4 and the mean is x ¯ , then x ¯ + x 6 is equal to :

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Step-by-step solution

More from Statistics

Let 9 = x 1 &#60; x 2 &#60; &#8230; &#60; x 7 be in an A.P. with common difference d . If the standard deviation of x 1 , x 2 &#8230; , x 7 is 4 and the mean is x &#175; , then x &#175; + x 6 is equal to :

Options

Correct answer

Step-by-step solution

More from Statistics

Let 9 = x 1 < x 2 < … < x 7 be in an A.P. with common difference d . If the standard deviation of x 1 , x 2 … , x 7 is 4 and the mean is x ¯ , then x ¯ + x 6 is equal to :