Answer:
Given:
The population mean $ (\mu) = 18 $
The population standard deviation $ (\sigma) = 2 $
Solution:
A) The $20^{th}$ percentile for incubation times:
$\therefore P(x < z) = 0.20 $
$\therefore z = -0.8416 $
→ Now, we can find the $20^{th}$ percentile.
$\text{Now, } z = \frac{x – \mu}{\sigma} $
$\therefore -0.8416 = \frac{x – 18}{2} $
$\therefore x = 16.317 $
$\approx 16 $
B) The incubation times that make up the middle 45%:
$\therefore \text{P}(-z < Z < z) = 0.45 $
$\therefore z = 0.5978 $
→ Now, we can find the middle 45%.
$\therefore \text{The middle } 45\% = \mu \pm z \times \sigma $
$ = 18 \pm 0.5978 \times 2 $
$ = (17, 19) $
Final Answer:
A) The $20^{th}$ percentile for incubation times $ \approx 16 $ days.
B) The incubation times that make up the middle 45% are between 17 and 19 days.
