# relation between location of a sample mean in sampling distribution and the standard error

Can someone explain the following statement with an example

We'll describe the location of the sample mean by calculating how many standard errors it is away from the center of the sampling distribution. That will give us a z-score for our sample mean.

Z-scores for test results. Suppose exam scores on a statewide math test are normally distributed with mean $$\mu = 300, \sigma =25.$$ A particular student scores $$X = 330,$$ what is her z-score?

$$Z = \frac{X - \mu}{\sigma} = \frac{330 - 300}{25} = \frac{30}{25} = 1.2.$$

She scores 30 points above the mean $$\mu,$$ which is $$30/25 = 1.2$$ standard deviations above the mean.

Suppose that last year's scores on an equivalent test were normally distributed with $$\mu = 305$$ and $$\sigma=27,$$ and that her older brother scored $$Y = 335.$$ Is it fair for him to brag that he's better at math than his sister? Aside from mentioning that bragging is futile and maybe nasty, we could point out that his z-score was only $$(335 - 305)/27 = 30/27 = 1.11.$$

One purpose of z-scores is to try to make results from slightly different situations more comparable. If his sister knows about z-scores, then he might decide to turn the conversation to one of his non-mathematical accomplishments.

Z-scores for standard errors. In your question you mention 'standard error'. That term is used in statistical inference. Suppose you have a sample of size $$n = 100$$ from a normal population with mean $$\mu = 50$$ and standard deviation $$\sigma = 5$$ (variance $$\sigma^2 = 25).$$ Also suppose the sample mean is $$\bar X = 50.8$$

Then the 'standard error of the mean' is $$\sigma_{\bar X} = SD(\bar X) = \frac{\sigma}{\sqrt{n}} = \frac{5}{10} = 0.5.$$ If you want to know how many standard errors the sample mean $$\bar X = 50.8$$ is above the population mean $$\mu = 50,$$ then you use the same formula as for test scores above: $$Z =\frac{\bar X - \mu}{SD(\bar X)} = \frac{\bar X - \mu}{\sigma/\sqrt{n}} = \frac{50.8-50}{0.5} = \frac{0.8}{0.5} = 1.6.$$

So the sample mean is 1.6 standard errors above the population mean. In many applications, such a z-score often lies in the interval $$(-2, 2)$$ when nothing remarkable has occurred. But I should leave the specific applications to your text and lectures.

Extra. More on z-scores for test results:

Another purpose of z-scores is to find various kinds of probabilities and proportions. We could use a printed table of the standard normal distribution to find the percentage of students who scored lower than the sister this year and the percentage who scored lower than her brother last year. These are called percentiles.

Maybe you can use the printed standard normal cumulative distribution table in your textbook to find the percentiles shown below. (I'm using R statistical software, but you should get similar percentiles, to about four significant digits, from the printed table.)

pnorm(1.2)
[1] 0.8849303     # sister's percentile is about 88.5
pnorm(1.11)
[1] 0.8665005     # brother's percentile is about 86.7

• Thanks a lot man!! This has been the best explanation to this question. I knew what is Z score and all this but now I know how it is used in real world. – Deshwal Oct 12 '19 at 13:11