Suppose **X** is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable,
suppose:

- µ
_{X}= the mean of X - σ
_{X}= the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable which consists of sample means, tends to be **normally distributed** and

**The Central Limit Theorem** for Sample Means says that if you keep drawing larger and larger samples (like rolling 1, 2, 5, and, finally, 10 dice) and **calculating their means** the sample means form their own **normal distribution** (the sampling distribution). The normal distribution has the same mean as the
original distribution and a variance that equals the original variance divided by **n**, the sample size. n is the number of values that are averaged together not the
number of times the experiment is done.

To put it more formally, if you draw random samples of size **n**, the distribution of the random variable , which consists of sample means, is called the **sampling distribution
of the mean**. The sampling distribution of the mean approaches a normal distribution as **n**, the sample size, increases.

The random variable has a different z-score associated with it
than the random variable **X**. is
the value of in one sample.

µ_{X} is both the average of X and of .

= standard deviation
of and is called the **standard error of
the mean.**

**Example 3.7**

An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population.

See Problem 1-2.

**Problem 1**

Find the probability that the **sample mean** is between 85 and 92.

**Solution**

Let X = one value from the original unknown population. The probability question asks you to find a probability for the **sample mean.**

Let = the mean of a sample of size 25. Since µ_{X} = 90,
σ_{X} = 15, and n = 25;

then

Find P (85 << 92) Draw a graph.

The probability that the sample mean is between 85 and 92 is 0.6997.

**TI-83 or 84**: `normalcdf`(lower value, upper value, mean, standard error of the mean)

The parameter list is abbreviated (lower value, upper value, µ, )

**Problem 2**

Find the value that is 2 standard deviations above the expected value (it is 90) of the sample mean.

**Solution**

To find the value that is 2 standard deviations above the expected value 90, use the formula

So, the value that is 2 standard deviations above the expected value is 96.

**Example 3.8**

The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a **mean of 2** hours and a **standard deviation of 0.5 hours**. A **sample of size n = 50** is drawn randomly from the population.

See Problem below.

**Problem**

Find the probability that the **sample mean** is between 1.8 hours and 2.3 hours.

**Solution**

Let X = the time, in hours, it takes to play one soccer match.

The probability question asks you to find a probability for the **sample mean time, in** **hours**, it takes to play one soccer
match.

Let the **mean** time , in hours, it
takes to play one soccer match.

If µ_{X} = ________ , σ_{X} = ________ , and n = ________ , then (________, ________) by the Central Limit Theorem for Means.

µ_{X} = 2, σ_{X} = 0.5, n = 50, and

Find . Draw a graph.

The probability that the mean time is between 1.8 hours and 2.3 hours is ________.

- 瀏覽次數：2208