Question

The lifetime of bacteria follows the Weibull distribution. The probability that the bacteria lives for more than 10 hours is 0.6 and that it lives more than 20 hours is 0.3. We want to find the probability that among 300 bacteria more than 200 live longer than 10 hours. Which of the following works best to compute this? O P(TW> 10; a and b to be computed from data) O PIXB > 200; p=0.6, n=300) O PIXB > 200.5: p=0.6, n=300) OPXN < 200: Mu = 180, Sigma = 180) The lifetime of bacteria follows the Weibull distribution. The probability that the bacteria lives for more than 10 hours is 0.65 and that it lives more than 20 hours is 0.3. The probability that among 300 bacteria more than 200 live longer than 10 hours can be computed as P(Z>a). What is the value of a? Please report your answer in 3 decimal places

Answer 1

option 2 is the correct choice to compute the probability that among 300 bacteria more than 200 live longer than 10 hours.

Step-by-step explanation:

To compute the probability that among 300 bacteria more than 200 live longer than 10 hours, we can use the binomial distribution.

The binomial distribution is used to model the number of successes in a fixed number of independent Bernoulli trials.

The formula for the binomial distribution is:P(X=k) = C(n,k) * p^k * (1-p)^(n-k)Where:- P(X=k) is the probability of getting exactly k successes.- n is the number of trials (300 bacteria in this case).- k is the number of successes (more than 200 bacteria live longer than 10 hours).- p is the probability of success in a single trial (probability that a bacterium lives longer than 10 hours).The answer options are:1.

In this case, we consider a success as a bacterium living longer than 10 hours, and the probability of success (p) is given as 0.6.

Therefore, option 2 is the correct choice to compute the probability that among 300 bacteria more than 200 live longer than 10 hours.