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This question was asked in a statistics session in my university:

The respective means of the first 8 and last 8 values of a statistical series of 15 values are 8 and 14. Given that the mean of the series is 12, then the 8th value is:

A. 2 B. 3 C. 4 D. 5 E. None of the above

The answer key says that (C) is the correct answer.

I tried in vain to solve this problem, but I couldn't figure out how.

I thought that since 8 is the mean of the first 8 values, therefore the 8th value is for sure greater than 8.

Can someone please explain? Any help or comment is appreciated. Thanks.

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Let's define your 15 values as $(x_{1},...,x_{15})$. We know that:

$$12 = \frac{1}{15}\sum_{i = 1}^{15} x_{i}$$

Now, we have that:

$$\frac{1}{15}\sum_{i = 1}^{15} x_{i} = \frac{1}{15} \left ( \sum_{i = 1}^{8} x_{i} + \sum_{i = 8}^{15} x_{i} - x_{8} \right ) $$

where $x_{8}$ is subtracted in the end since otherwise you would be summing it two times, as last term in the first summation and first term in the second summation.

But since we are also told that:

$$\frac{1}{8}\sum_{i = 1}^{8} x_{i} = 8 \Rightarrow \sum_{i = 1}^{8} x_{i} = 64$$

and that:

$$\frac{1}{8}\sum_{i = 8}^{15} x_{i} = 14 \Rightarrow \sum_{i = 8}^{15} x_{i} = 112$$

we can conclude that:

$$12 = \frac{1}{15} \left ( 64 + 112 - x_{8} \right )$$

leading to $$x_{8} = -4$$

Possibly a typo in your question?

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    $\begingroup$ Your answer seems logical; it is possibly a typo in the sessions book. Thank you. $\endgroup$ – user208973 Mar 28 at 13:12

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