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Every exam has 100 points. Student's average of 3 exams is 88 points and standard deviation is 3. How many points should he get on fourth and fifth exam so that average is 90 and standard deviation stays same (=3).

I tried solving equations $(x_1+x_2+x_3)/3=88$ and $(x_1+x_2+x_3+x_4+x_5)/5=90$. Since $x_1+x_2+x_3=264$ it means that $(264+x_4+x_5)/5=90$ so $x_4+x_5=186$. Then I tried to use equations for standard deviation but I haven't managed to get anything useful.

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  • $\begingroup$ What if I told you there is no solution? $\endgroup$ – Parcly Taxel Jan 31 '18 at 11:36
  • $\begingroup$ How to proof that? $\endgroup$ – Mirjan Pecenko Jan 31 '18 at 11:41
  • $\begingroup$ See my answer.${}$ $\endgroup$ – Parcly Taxel Jan 31 '18 at 11:49
  • $\begingroup$ Is the std defined by the population or sample formula ? $\endgroup$ – Yves Daoust Jan 31 '18 at 11:51
  • $\begingroup$ I think by the population $\endgroup$ – Mirjan Pecenko Jan 31 '18 at 11:54
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We have $$\sqrt{\frac{(x_1-88)^2+(x_2-88)^2+(x_3-88)^2}3}=3$$ $$(x_1-88)^2+(x_2-88)^2+(x_3-88)^2=27$$ How do we get from $(x-88)^2=x^2-176x+88^2$ to $(x-90)^2=x^2-180x+90^2$? We add $-4x+356$. $$-4x_1+356-4x_2+356-4x_3+356=-4(x_1+x_2+x_3)+1068=-4(3\cdot88)+1068=12$$ $$(x_1-90)^2+(x_2-90)^2+(x_3-90)^2=27+12=39$$ Now $$\sqrt{\frac{39+(x_4-90)^2+(x_5-90)^2}5}=3$$ $$(x_4-90)^2+(x_5-90)^2=6$$ Let $z_4=x_4-90$ and $z_5=x_5-90$. Then $z_4+z_5=6$ while $z_4^2+z_5^2=6$. Squaring the first equation and subtracting the second from it and halving gives $z_4z_5=15$.

Thus we may form the polynomial $x^2-6x+15$, whose roots should be $z_4$ and $z_5$. However, we find that the polynomial has no real roots. So there is no solution to the original problem – the average and standard deviation cannot be changed by the further two tests to the given values.

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From the averages,

$$x_4+x_5=5\cdot 90-3\cdot 88=186$$

and from the variances,

$$x_4^2+x_5^2=5(3^2+90^2)-3(3^2+88^2)=17286.$$

Then

$$(x_4-x_5)^2=2(x_4^2+x_5^2)-(x_4+x_5)^2=-24.$$ (!)

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