[Math] Simple question on standard deviation and mean.

standard deviationstatistics

This question has kinda stumped me:

For a data set, the mean is 5 and the standard deviation is 1. There
are 10 values in the data set. Is the range of the data set bigger,
smaller or equal to 10?

I don't want the answer, just the technique. Here's what I've tried (admittedly, not much):

$$\sum_{i=0}^{9}a_i / 10= 5$$
$$\sigma = 1$$
Is there a relation between standard deviation and simple mean which I can use? Any help is appreciated. This is not homework, it's a practice qn for the GRE.

Best Answer

$\def\implies{\quad\Rightarrow\quad}$If the mean is 5, then you know that

$$\frac{1}{10} \sum_i a_i = 5 \implies \sum_ia_i = 50$$

and if the standard deviation is 1 (and hence the variance is 1) then

$$\frac{1}{10} \sum_i(a_i-5)^2 = 1 \implies \sum_i(a_i-5)^2 = 10$$

To make the range as big as possible, we choose two values $a^+$ and $a^-$ to be as far from the mean as possible (symmetrically) and set the rest of the values to the mean, i.e. $a^+=5+x$ and $a^-=5-x$. What does that tell you about $x$?

Related Solutions

[Math] Calculating mean and standard deviation of very large sample sizes

Posting as an answer in response to comments.

Here's a way to compute the mean and standard deviation in one pass over the file. (Pseudocode.)

n = r1 = r2 = 0;
while (more_samples()) {
    s = next_sample();
    n += 1;
    r1 += s;
    r2 += s*s;
}
mean = r1 / n;
stddev = sqrt(r2/n - (mean * mean));

Essentially, you keep a running total of the sum of the samples and the sum of their squares. This lets you easily compute the standard deviation at the end.

[Math] Statistics: how does standard deviation measure quality – please answer everyone

There are at least two senses in which you can use the standard deviation in quality measurement:

Manufacturing precision: How closely are we able to meet a manufacturing specificiation (e.g., bearing diameter).
Fraction falling within a given tolerance interval: Continuing the example, if bearings must be within $\pm 1$ mm, how many standard deviations does this represent GIVEN our observed manufacturing variability (i.e., (1) from above).

The more fundamental use of the standard deviation is in (1), where you are characterizing how well-controlled your manufacturing process is. In this case, the larger the standard deviation is, the lower the quality of your manufacturing process. This is regardless of the actual standards you need to meet - if process A has higher standard deviation than process B, than for any tolerance interval $\pm a$ mm, process A will generate more bad products than process B.

Now, the second use of standard deviation, (2) from above, is probably what is causing you confusion. In this case, you are correct that more standard deviations indicate a higher quality process, but these standard devaitions are NOT the same as in (1). To illustrate, imagine that you have a process that produces bearings with mean diameter 0.2 mm with standard deviation 0.01 mm (and normally distributed). Now, your tolerance interval is $\pm .05$ mm. How many standard deviations does $\pm .05$ represent, given that you observed a process standard deviation of 0.01 mm? In this case, you would say that your process is $5\sigma$, in that the predetermined tolerance limits represent 5 standard deviations away from your mean process output (which I assumed to be unbiased), as calculated from the underlying process data.

So, you see, you want the standard devation of your manufacturing process to be low, which will increase the number of standard deviations that can fit in your pre-specified tolerance interval. The two uses of $\sigma$ do not mean the same thing.

Best Answer

Related Solutions

[Math] Calculating mean and standard deviation of very large sample sizes

[Math] Statistics: how does standard deviation measure quality – please answer everyone

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