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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:
If you draw random samples of size n , then as n increases, the random variable $\overline{X}$ which consists of sample means, tends to be normally distributed and
$\overline{X}$ ~ N $\left({\mu}_{x}\text{,}\frac{\sigma x}{\sqrt{n}}\right)$ .
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten 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, the sample size. The variable 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 $\overline{X}$ , 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 $\overline{X}$ has a different z -score associated with it from that of the random variable X . The mean $\overline{x}$ is the value of $\overline{X}$ in one sample.
μ _{ X } is the average of both X and $\overline{X}$ .
$\sigma \overline{x}\text{=}\frac{\sigma x}{\sqrt{n}}$ = standard deviation of $\overline{X}$ and is called the standard error of the mean.
To find probabilities for means on the calculator, follow these steps.
2nd DISTR
2:normalcdf
$normalcdf\left(lower\text{}value\text{}of\text{}the\text{}area,\text{}upper\text{}value\text{}of\text{}the\text{}area,\text{}mean,\text{}\frac{standarddeviation}{\sqrt{samplesize}}\right)$
where:
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.
a. Find the probability that the
sample mean is between 85 and 92.
a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean .
Let $\overline{X}$ = the mean of a sample of size 25. Since μ _{ X } = 90, σ _{X} = 15, and n = 25,
$\overline{X}$ ~ N $\left(90\text{,}\frac{15}{\sqrt{25}}\right)$ .
Find P (85< $\overline{x}$ <92). Draw a graph.
P (85< $\overline{x}$ <92) = 0.6997
The probability that the sample mean is between 85 and 92 is 0.6997.
normalcdf
(lower value, upper value, mean, standard error of the mean)
The parameter list is abbreviated (lower value, upper value, μ , $\frac{\sigma}{\sqrt{n}}$ )
normalcdf
(85,92,90,
$\frac{15}{\sqrt{25}}$ ) = 0.6997
b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.
b. To find the value that is two standard deviations above the expected value 90, use the formula:
value = μ _{x} + (#ofTSDEVs) $\left(\frac{{\sigma}_{x}}{\sqrt{n}}\right)$
value = 90 + 2 $\left(\frac{15}{\sqrt{25}}\right)$ = 96
The value that is two standard deviations above the expected value is 96.
The standard error of the mean is $\frac{\sigma x}{\sqrt{n}}$ = $\frac{15}{\sqrt{25}}$ = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n .
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