6.2 The central limit theorem for sample means (averages)  (Page 7/5)

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 ${\mu }_X=90$ , ${\sigma }_X=15$ , and $n_{}=25$ ;

then $\overline{X}$ ~ $N(90,\frac{15}{\sqrt{25}})$

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.

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

The parameter list is abbreviated (lower value, upper value, $\mu$ , $\frac{\sigma }{\sqrt{n}}$ )

normalcdf $\mathrm{(85,92,90,}\frac{15}{\sqrt{25}}\mathrm{)\; =\; 0.6997}$

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

value = ${\mu }_X+\text{(\#ofSTDEVs)}\left(\frac{{\sigma }_X}{\sqrt{n}}\right)$

value = $90+2\cdot \frac{15}{\sqrt{25}}=96$

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

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 ${\mu }_X=$ _________, ${\sigma }_X=$ __________, and $n=$ ___________, then $\overline{X}~N\text{(\_\_\_\_\_\_, \_\_\_\_\_\_)}$ by the Central Limit Theorem for Means.

${\mu }_X=$ 2 , ${\sigma }_X=$ 0.5 , $n=$ 50 , and $X~N(2,\frac{0.5}{\sqrt{50}})$

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

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

normalcdf $\mathrm{(1.8,2.3,2,}\frac{.5}{\sqrt{50}}\mathrm{)\; =\; 0.9977}$

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