# Probability with samples?

## Answer whichever part (a-e) you can/want to :)

Jul 25, 2017

a) $P \left(x \setminus < 40\right) \setminus \approx 0.189$

b) $P \left(\setminus \overline{x} \setminus < 40\right) , n = 9 \setminus \Rightarrow \setminus \approx 0.0041$

c) $P \left(\setminus \overline{x} \setminus < 40\right) , n = 12 \setminus \Rightarrow \setminus \approx 0.0011$

d) not sure see explanation for more

e) unusual as it exceeds 2 z-scores (or 2 standard deviations)

#### Explanation:

Given information
$\setminus \mu = 43.7$ cm, $\setminus \sigma = 4.2$ cm

Working it out

a) $P \left(X \setminus < 40\right) = P \left(z \setminus < \frac{40 - 43.7}{4.2}\right) = P \left(z \setminus < - \frac{3.7}{4.2}\right)$$= P \left(z \setminus < - 0.8810\right) \setminus \approx 0.189$

b) $n = 9 \setminus \rightarrow \setminus {\sigma}_{\setminus \overline{x}} = \frac{4.2}{\setminus} \sqrt{9} = 4.2 \setminus \div 3 = 1.4 \setminus \Rightarrow P \left(\setminus \overline{x} \setminus < 40\right)$$= P \left(z \setminus < - 3.7 \setminus \div 1.4\right) = P \left(z \setminus < - 2.6429\right) \setminus \approx 0.0041$

c) $n = 12 \setminus \rightarrow \setminus {\sigma}_{\setminus \overline{x}} = \frac{4.2}{\setminus} \sqrt{12} \setminus \approx 1.2124 \setminus \Rightarrow P \left(\setminus \overline{x} \setminus < 40\right)$$= P \left(z \setminus < - 3.7 \setminus \div 1.2124\right) = P \left(z \setminus < - 3.0517\right) \setminus \approx 0.0011$

d) Increasing sample size appears to lower probability. This is likely because --

e) $n = 15 \setminus \rightarrow \setminus {\sigma}_{\setminus \overline{x}} = \frac{4.2}{\setminus} \sqrt{15} \setminus \approx 1.0844$\bar{x}=46;z=(46-43.7)/(1.0844)\approx2.1209
This result is unusual as it exceeds a z-score of 2.