- (a) Determine the 15th percentile for incubation times.
- (b) Determine the incubation times that make up the middle 39%.
(Round to the nearest whole number as needed.)
Answer:
Given:
The population mean $(\mu) = 21 $
The population standard deviation $(\sigma) = 1$
Solution:
A) The $ 15^(th) $ percentile for incubation times:
$\therefore P(x < z) = 0.15 $
$ therefore z = -1.036 $
→ Now, we can find $15^(th)$ percentile.
$\text{Now, } z = \frac{x – \mu}{\sigma} \ $
$ \therefore -1.036 = \frac{x – 21}{1} \ $
$ \therefore x = 19.964 \ $
$ \approx 20 $
B) The incubation times that make up the middle 39%:
$ \therefore \text{P}(-z < z < z) = 0.39 \ $
$ \therefore z = 0.51 $
→ Now, we can find the middle 39%.
$ \therefore \text{The middle } 39\% = \mu \pm z \times \sigma $
$ = 21 \pm 0.51 \times 1 $
$ = (20, 22) $
Final Answer:
A) The $15^{th}$ percentile for incubation times $ \approx 20 $ days.
B) The incubation times that make up the middle 39% are between 20 and 22 days
