Question

Suppose the number of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water tend to be approximately normally distributed, with a mean of 85 and a standard deviation of 9. What is the probability that a given 1-ml will contain more than 100 bacteria

Answer 1

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: $\mu=85,\ \ \sigma=9$

The probability that a given 1-ml will contain more than 100 bacteria will be:

$P(X>100)=P(\dfrac{X-\mu}{\sigma}>\dfrac{100-85}{9})\\\\=P(Z>1.67)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Zz)=1-P(Z$

∴The probability that a given 1-ml will contain more than 100 bacteria

0.0475.