Answer: 0.0475
Step-by-step explanation:
Let x = random variable that represents the number of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water, such that X is normally distributed.
Given: [tex]\mu=85,\ \ \sigma=9[/tex]
The probability that a given 1-ml will contain more than 100 bacteria will be:
[tex]P(X>100)=P(\dfrac{X-\mu}{\sigma}>\dfrac{100-85}{9})\\\\=P(Z>1.67)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z<1.67)\ \ \ [P(Z>z)=1-P(Z<z)]\\\\=1- 0.9525=0.0475[/tex]
∴The probability that a given 1-ml will contain more than 100 bacteria
0.0475.
