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: 
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](https://tex.z-dn.net/?f=P%28X%3E100%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B100-85%7D%7B9%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.67%29%5C%20%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C1.67%29%5C%20%5C%20%5C%20%5BP%28Z%3Ez%29%3D1-P%28Z%3Cz%29%5D%5C%5C%5C%5C%3D1-%200.9525%3D0.0475)
∴The probability that a given 1-ml will contain more than 100 bacteria
0.0475.
Answer:
2.5-5
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
5-1=4
Answer: 3.97000947 × 10^4