If the standard deviation of the numbers 2, 3 | vedclass

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(A, B) Given the numbers are $2, 3, 2x, 11$. The mean $\bar{x} = \frac{2+3+2x+11}{4} = \frac{16+2x}{4} = 4 + \frac{x}{2}$.Given standard deviation $\s

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$7/2$

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(A, B) Given the numbers are $2, 3, 2x, 11$. The mean $\bar{x} = \frac{2+3+2x+11}{4} = \frac{16+2x}{4} = 4 + \frac{x}{2}$.Given standard deviation $\sigma = 3.5 = \frac{7}{2}$,so $\sigma^2 = \frac{49}{4}$.The variance formula is $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.$\frac{49}{4} = \frac{2^2 + 3^2 + (2x)^2 + 11^2}{4} - (4 + \frac{x}{2})^2$.$\frac{49}{4} = \frac{4 + 9 + 4x^2 + 121}{4} - (16 + 4x + \frac{x^2}{4})$.$\frac{49}{4} = \frac{134 + 4x^2}{4} - \frac{64 + 16x + x^2}{4}$.$49 = 134 + 4x^2 - 64 - 16x - x^2$.$49 = 3x^2 - 16x + 70$.$3x^2 - 16x + 21 = 0$.$(3x - 7)(x - 3) = 0$.Thus,$x = \frac{7}{3}$ or $x = 3$.

If the standard deviation of the numbers $2, 3, 2x$,and $11$ is $3.5$,find the possible values of $x$.

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