Answer:
The probability of a pill having more than 24.125 mg active substance is 0.0495.
Explanation:
Let X denote the amount of active substance in each pill.
It is provided that,
![X\sim N(20,\ 2.5^(2))](https://img.qammunity.org/2021/formulas/mathematics/college/vzzqwpdu3ojqplnud7u1mjsaiah15n4ly0.png)
Compute the probability of a pill having more than 24.125 mg active substance as follows:
![P(X>24.125)=P((X-\mu)/(\sigma)>(24.125-20)/(2.5))](https://img.qammunity.org/2021/formulas/mathematics/college/7pjckpvawez7mwqe2wievc8psw9aze4wlz.png)
![=P(Z>1.65)\\\\=1-P(Z<1.65)\\\\=1-0.95053\\\\=0.04947\\\\\approx 0.0495](https://img.qammunity.org/2021/formulas/mathematics/college/dbfz96ago8mcl5w1oot5bjv7x39swz2ppr.png)
Thus, the probability of a pill having more than 24.125 mg active substance is 0.0495.