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Mathematics

The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15, find the 9th variate.

Measures of Central Tendency

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Answer

Let 9th variate be x.

Sum of terms = 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 + x = 89 + x

Number of terms = 9.

Arithmetic mean (A.M.) = Sum of termsNo. of terms\dfrac{\text{Sum of terms}}{\text{No. of terms}}

Given, A.M. = 11

∴11=89+x9⇒99=89+x⇒x=99−89⇒x=10.\therefore 11 = \dfrac{89 + x}{9} \\[1em] \Rightarrow 99 = 89 + x \\[1em] \Rightarrow x = 99 - 89 \\[1em] \Rightarrow x = 10.

Hence, the value of 9th variate is 10.

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