Answer:
Given:
The population mean $ (\mu) = 30 $
The population standard deviation $ (\sigma) = 1.5 $
Solution:
A) The $25^{th}$ percentile for incubation times:
$\therefore P(x < z) = 0.25 $
$\therefore z = -0.6745 $
→ Now, we can find the $25^{th}$ percentile.
$\text{Now, } z = \frac{x – \mu}{\sigma} $
$\therefore -0.6745 = \frac{x – 30}{1.5} $
$\therefore x = 28.988 $
$\approx 29 $
B) The incubation times that make up the middle 50%:
$\therefore \text{P}(-z < Z < z) = 0.50 $
$\therefore z = 0.6745 $
→ Now, we can find the middle 50%.
$\therefore \text{The middle } 50\% = \mu \pm z \times \sigma $
$ = 30 \pm 0.6745 \times 1.5 $
$ = (29, 31) $
Final Answer:
A) The $25^{th}$ percentile for incubation times $ \approx 29 $ days.
B) The incubation times that make up the middle 50% are between 29 and 31 days.
