Answer:
Given:
The population mean $ (\mu) = 50 $
The population standard deviation $ (\sigma) = 3 $
Solution:
A) The $10^{th}$ percentile for incubation times:
$\therefore P(x < z) = 0.10 $
$\therefore z = -1.2816 $
→ Now, we can find the $10^{th}$ percentile.
$\text{Now, } z = \frac{x – \mu}{\sigma} $
$\therefore -1.2816 = \frac{x – 50}{3} $
$\therefore x = 46.155 $
$\approx 46 $
B) The incubation times that make up the middle 30%:
$\therefore \text{P}(-z < Z < z) = 0.30 $
$\therefore z = 0.3853 $
→ Now, we can find the middle 30%.
$\therefore \text{The middle } 30\% = \mu \pm z \times \sigma $
$ = 50 \pm 0.3853 \times 3 $
$ = (49, 51) $
Final Answer:
A) The $10^{th}$ percentile for incubation times $ \approx 46 $ days.
B) The incubation times that make up the middle 30% are between 49 and 51 days.
