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The question that i'm being asked it: the mean of 10 observed scores was 20 and the standard deviation if 6.0. the observed scores are {16,11,20,24,29,24,16,20,,} what are the two missing scores

I have tried to figure this out and i keep ending up needing to use the quadratic formula and getting two numbers that equate to a different standard deviation. I really do not know what more i can do. please help me

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Do you use $n$ or $n-1$ in calculations of standard deviation? (Reason I am asking is that there is not uniformity of usage.) –  André Nicolas Oct 10 '13 at 2:40
    
Our class has been using n-1 for it –  user99945 Oct 10 '13 at 2:41

2 Answers 2

up vote 1 down vote accepted

Let the missing scores be $a$ and $b$. The non-missing scores have mean $20$, so the missing scores must add up to $40$.

The standard deviation, if we use "$n-1$", not $n$, is equal to $$\sqrt{\frac{1}{9} (226+(a-20)^2+(b-20)^2)}.$$ This is equal to $6$, so some manipulation gives $$324=226+(a-20)^2+(b-20)^2.$$ Let $x=a-20$ and $y=b-20$. We get $x+y=0$ and $x^2+y^2=98$. Substitute. We get $x^2=49$. So $x=7$ and $y=-7$, or the other way around. Our missing values are $13$ and $27$.

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Thank you so much :) –  user99945 Oct 10 '13 at 3:10
    
You are welcome. You probably got near this, except for a minor glitch. –  André Nicolas Oct 10 '13 at 3:17

It sounds as if you were probably approaching the problem sensibly. If the missing scores are $x$ and $y$, we know that $160+x+y=10\cdot20=200$, so $x+y=40$. We also know that $$6^2=\frac1{10}\sum_{k=1}^{10}s_k^2-20^2\;,\tag{1}$$ where the $s_k$ are the scores, so $4360=3426+x^2+y^2$, and $x^2+y^2=934$. Set $y=40-x$ and substitute:

$$x^2+(40-x)^2=934\;,$$

so $2x^2-80x+1600=934$, $2x^2-80x+666=0$, and $x^2-40x+333=0$. Then

$$x=\frac{40\pm\sqrt{268}}2=20\pm2\sqrt{67}\;,$$

making the two scores approximately $36.37$ and $4.63$, which are unexpectedly messy.

If $6.0$ is supposed to be a sample standard deviation, however, $(1)$ becomes

$$6^2=\frac{10}9\left(\frac1{10}\sum_{k=1}^{10}s_k^2-20^2\right)\;,$$

which reduces to $4324=3426+x^2+y^2$, or $x^2+y^2=898$. Then we get $x^2-40x+351=0$, and

$$x=\frac{40\pm\sqrt{196}}2=20\pm7\;,$$

and the missing scores turn out to be $13$ and $27$. This is probably the intended solution.

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