6.1 The central limit theorem for sample means (averages) -- rrc

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.

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 .

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 $\overline{X}$ = the mean time, in hours, it takes to play one soccer match.

If μ _X = _________, σ _X = __________, and n = ___________, then X ~ N (______, ______) by the central limit theorem for means .

μ _X = 2, σ _X = 0.5, n = 50, and X ~ N $\left(\text{2, }\frac{0.5}{\sqrt{50}}\right)$

Find P (1.8< $\overline{x}$ <2.3). Draw a graph.

P (1.8< $\overline{x}$ <2.3) = 0.9977

The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.