See that $\overline{x}$ is a random variable with its own distribution

State the Central Limit Theorem: center, spread, shape

Compute probabilities and percentiles for a sample mean

Different samples from the same population will generally produce different values of $\overline{x}$. Because $\overline{x}$ varies from sample to sample, it is a random variable with its own probability distribution. This distribution is called the sampling distribution of $\overline{x}$.

Go to humanbenchmark.com and take the Verbal Memory test.

Words flash on the screen one at a time.

For each word, click "SEEN" if you've seen it before in this session, or "NEW" if it's the first time.

Your score is the number of correct answers before three mistakes.

Nothing to report and nothing to hand in. Just go see what it is.

That class averaged 40.5. Typical for a class of 36? Or unusually low?

That's a probability question!

We need to know what the distribution of sample means looks like when samples of this size are drawn from the population.

Draw more samples

The sampling distribution of $\overline{x}$ is approximately normal, even though the population is skewed right.

The population mean is 52.

The mean of the sample means lands right next to it. On average, the sample mean equals the population mean: the sampling distribution of $\overline{x}$ is centered on $\mu$.

The population standard deviation is 34.6.

The standard deviation of the sample means is far smaller. Averaging 36 scores cancels the highs against the lows, so the sample means are packed in much more tightly than individual scores are.

$\frac{34.6}{\sqrt{36}}=\frac{34.6}{6}=5.77$

${\sigma }_{\overline{x}}$ is called the standard error of the mean.

The machine's number is close to 5.77, and it gets closer with more samples.

Same skewed population. Samples of $n=5$.

Same skewed population. Samples of $n=10$.

Same skewed population. Samples of $n=30$.

The Central Limit Theorem applies to all populations.

A fair die: 1, 2, 3, 4, 5, 6. Every face equally likely. Uniform, and symmetric.

A normal population.

If NEITHER holds, we cannot use the Central Limit Theorem to find probabilities.

Check the two conditions: a large sample, or a normal population.

(a) A simple random sample of size 45 will be drawn from a population with mean $\mu =15$ and standard deviation $\sigma =3.5$.

(b) A simple random sample of size 24 will be drawn from a population with mean $\mu =35$ and standard deviation $\sigma =1.2$.

(c) A simple random sample of size 8 will be drawn from a normal population with mean $\mu =-60$ and standard deviation $\sigma =5$.

Check that $n>30$ (or the population is normal)

Find ${\mu }_{\overline{x}}=\mu$ and the standard error ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Same normal curve work as the last section: an area under the curve, or the value that goes with an area

What is the probability that the sample mean age is greater than 26 years?

Does the Central Limit Theorem apply? Then find the mean and standard error of $\overline{x}$ before touching the calculator.

$\mu =25$ years

$\sigma =9.5$ years

$n=125$

Find the 30th percentile of the sample mean age.

A percentile is a value of $\overline{x}$, not a probability.

$\mu =25$ years

$\sigma =9.5$ years

$n=125$

We know:

The 30th percentile is the value with area 0.30 to its left.

$\mu =52$ points

$\sigma =34.6$ points

$n=36$

$\overline{x}=40.5$

What is the probability that the sample mean is less than 70 minutes? Would this be unusual?

Two parts: compute the probability, then judge it. “Unusual” means a probability below 0.05.

$\mu =87$ minutes

$\sigma =45$ minutes

$n=50$

Above what value would only 5% of sample means fall?

“Only 5% above” means area 0.95 to the left of the cutoff, and the answer is a value of $\overline{x}$, not a probability.

$\mu =34$ minutes

$\sigma =14$ minutes

$n=49$

Only 5% above the cutoff means area 0.95 to its left.

Suppose a large airport says the average wait at its main security checkpoint, on weekday mornings from 8:00 to 8:30, is 18 minutes. Waits are skewed right, with standard deviation 12 minutes.

A local newspaper thinks the airport's posted waits run low. Does the evidence show the average is more than 18 minutes?

$\mu =18$ minutes

$\sigma =12$ minutes

$n=36$

The reporter goes back and times 36 travelers the same way: $\overline{x}=22$ minutes.

$n=36>30$, so the Central Limit Theorem applies.

Only 5% above the cutoff means area 0.95 to its left.

Is 21.29 also the 95th percentile of one wait?

No. One wait has standard deviation 12, not 2.