15 Day 15

Non-standard Normal Distributions

If \(X\) is a normal random variable with mean \(\mu\) and standard deviation \(\sigma\) we write \(X \sim N(\mu,\sigma^2)\)

  • \(\sim\) means “is distributed (as)”

  • \(N()\) refers to the normal distribution

    • Together we’re saying “X is distributed normal with mean \(\mu\) and variance \(\sigma^2\)

  • So if \(\mu=100\) and \(\sigma=5\) we’ll write \(X\sim (100,5^2)\) or \(X \sim (100,25)\)

  • If we write \(X \sim (16,5)\) then \(\mu=16\) and \(\sigma=\sqrt{5}\)


We’ve seen that the standard normal distribution is well understood and we can find probabilities, percentiles, etc. “easily” using a \(z\)-table

If we want to easily learn these things about non-standard random variables it would be convenient if we could transform them into standard normal random variables

  • Fortunately we can

\[\text{if} \ X \sim N(\mu,\sigma^2), \ \text{then} \ Z={X-\mu \over \sigma} \sim N(0,1)\]

\[X \sim N(100,10) \Rightarrow Z={X-100 \over 10} \sim N(0,1)\]


\[z={x-\mu \over \sigma}\]

What we’ve done is convert the non-standard normal distribution into a set of \(z\)-scores

  • What do these \(z\)-scores measure?


The process we’ve performed in making this conversion is called standardizing a normal random variable

  • We can do this moving forward to find out information from a non-standard normal r.v. using the \(z\)-tables

\[P(X\leq x)=P\left({X-\mu \over \sigma}\leq{x-\mu \over \sigma}\right)=P(Z\leq z)\]

Non-standard Normal Examples

Suppose that the heights of American men (\(20\) years and older) are approximately normal with a mean of \(70\) inches and a standard deviation of \(4\) inches.

  1. What proportion of American men are less than \(6\) feet tall?

  • (\(6\)’ \(=\) \(72\)”)

  • \(X \sim N(70,4^2)\)

\[P(X\leq 72)=P\left(Z\leq{72-70 \over 4}\right)=P(Z\leq 0.5)\]

  1. What proportion of American men are between 5’ and 6’ 8”tall?
  • (\(5\)’\(=\) \(60\)” and \(6\)’\(8\)” \(=\) \(80\)”)

\[P(60<X<80)=P\left({60-70 \over 4}<Z<{80-70 \over 4}\right)\]

\[=P(-2.50<Z<2.50) =2\times P(Z<-2.50)\]

\[=1-2(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ )=\]



Given our height example (\(X \sim N(70,4^2)\)), how tall would you have to be so that you are taller than \(90\%\) of American men?

  • Refer to your \(z\)-table

    • Find the closest value to \(0.90\)

    • “un-standardize” the value:

\[x=\sigma z + \mu=4( \ \ \ \ \ \ \ \ \ \ \ \ \ ) + 70 =\]


What is an interval of two heights that contains approximately \(50\%\) of American men?

  • Find a value on your \(z\)-table that is approximately \(0.25\) or \(0.75\)

    • Why these values?

  • We should end up with \(0.67\) at \(0.75\)

    • at \(0.25\) it should be \(-0.67\)

\[x=\sigma (-z) + \mu=4(-0.67) + 70 =67.32\]

\[x=\sigma z + \mu=4(0.67) + 70 =72.68\]



Sampling Distribution of Sample Mean & Central Limit

Let’s remember some core vocabulary:

  • Population: The entire collection of individuals we’re seeking information from

  • Sample: A subset of a population of which we can gather real observations from

  • Parameter: A value derived from a population

  • Statistic: A value derived from a sample


Realistically we will never quantify a parameter directly from a population

  • The major goal of the statistical sciences is to make inference about a population and its parameters by gathering a sample and deriving statistics

In practice:

  • Start with a research question

    • “How effective are seasonal Influenza vaccine campaigns in Kansas?”

\[ \begin{array}{|c|c|c|c|} \hline Population & Parameter & Sample & Statistic \\ \hline \text{Kansas Residents} & p_V & 10 \ \text{Kansas Towns} & \hat{p}_V\\ \hline \\ \hline \\ \hline \\ \hline \end{array} \]



Business Week reported on the cost per treatment of Herceptin, a drug used to treat breast cancer. Typical treatment costs (in dollars) for Herceptin are provided by a simple random sample of 5 patients.

\[ \begin{array}{|c|c|c|c|c|} \hline 4376 & 5578 & 2717 & 4920 & 4495 \\ \hline \end{array} \]

Find a number that can be used as an estimate of the mean cost per treatment with Herceptin.


  • Suppose we are interested in determining the average time (in minutes) it takes K-State students to travel to their hometowns.

  • We take a simple random sample of 100 K-State students, ask each selected student how long it takes to travel home, and then compute the sample mean:

\[\bar{x} = 91.34\]

  • Suppose we take another sample of 100 K-State students. This time our sample mean is:

\[\bar{x} = 89.63\]

  • If we view taking a random sample as an experiment, then the sample mean \(\bar{x}\) is a numerical value assigned to each outcome of the experiment.


We’ve discussed this previously, \(\bar{x}\), our sample mean, is a random variable

  • When our value is arising from a sample, a limited subset of the population, it’s value with vary each time our sample changes

  • So all statistics derived from a sample are random variables


This is a fundamental concept to grasp for all of statistics:

  • All random variables have a random probability distribution

    • As all statistics are random variables:

      • All statistics arise from a random probability distribution


We refer to the probability distribution of a sample statistic as the sampling distribution

  • We’re going to look at this through the lense of the sample mean


Let \(\bar{x}\) be the mean of a random sample of size \(n\), drawn from a population with mean \(\mu\) and standard deviation \(\sigma\)

Since \(\bar{x}\) is a random variable, it has the mean and the standard deviation

  • The mean of \(\bar{x}\) is \(\mu\). That is,

\[\mu_{\bar{x}} = \mu = \text{population mean}\]

  • The standard deviation of \(\bar{x}\) is \(\sigma / \sqrt{n}\). That is,

\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{\text{population std. deviation}}{\sqrt{\text{sample size}}}\]


a. A population has mean \(\mu = 6\) and standard deviation \(\sigma = 4\). Find \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) for a sample size of \(n = 25\)



b. A population has mean \(\mu = 17\) and standard deviation \(\sigma = 20\). Find \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) for a sample size of \(n = 100\)



The mean and standard deviation of the sample mean \(\bar{x}\) are

\[\mu_{\bar{x}} = \mu\]

\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]

  • This is true even when the true values of \(\mu\) and \(\sigma\) are unknown

  • This is how we make inference about population parameters with only sample statistics

  • We know the values of two parameters associated with the sampling distribution of \(\bar{x}\)

    • To fully understand its distribution, we also need to know its shape

    • Accessing all of this information is typically done through something called an exploratory analysis