Mean Absolute Deviation

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New answer posted•a year ago

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3},$  respectively, then the value of $| a - b |$  is equal to

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a year ago

10 = $\frac{7 + 10 + 11 + 15 + a + b}{6} \Rightarrow a + b = 17$ . (i)

$\Rightarrow \frac{20}{3} = \frac{7^{2} + 1 0^{2} + 1 1^{2} + 1 5^{2} + a^{2} + b^{2}}{6} - 1 0^{2}$

a² + b² = 145 . (ii)

solving (i) and (ii) we get a = 8, b = 9 or a = 9, b = 8

|a – b|= 1

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New answer posted•a year ago

2 Views|Posted a year ago

Answered by

a year ago

a, b, c, d, e be 5 unknown

n = 7, mean = 8, variance = 16

sum of observations = 7 × 8 = 56

mean of 5 remaining observation = $\frac{56 - 8 - 6}{25} = \frac{42}{5}$

$16 = \frac{\sum x_{i}^{2}}{7} - 64$

$\sum x_{i}^{2} = 560$

$\begin{array}{l} a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 460 \\ = 460 - {(\frac{42}{5})}^{2} \end{array}$

$= \frac{536}{25}$ &nbs

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