Jayla starts with a population of 100 amoebas that increases 40% every hour for a number of hours, h. The expression 100(1 + 0.4
2 answers:
You might be interested in
Answer:
a. P(X ≤ 5) = 0.999
b. P(X > λ+λ) = P(X > 2) = 0.080
Step-by-step explanation:
We model this randome variable with a Poisson distribution, with parameter λ=1.
We have to calculate, using this distribution, P(X ≤ 5).
The probability of k pipeline failures can be calculated with the following equation:
Then, we can calculate P(X ≤ 5) as:
The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:
Answer:
Probability = 0.502
Step-by-step explanation:
We are given the following data :
Hours Count Percent
1 18 3.44
2 55 10.50
3 81 15.46
4 109 20.80
5 88 16.79
6 66 12.60
7 39 7.44
8 17 3.24
9 17 3.24
10 19 3.63
15 15 2.86
We need to calculate the probability
P(Length of stay of exactly 1 is less than or equal to 4)
P() = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)
We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.