Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 48 days and a standard deviation of 10.5 days. Find the probability that a simple random sample of 36 protozoa will have a mean life expectancy of 51 or more days.

P (x_bar ≥ 51) = 0.0432

Step-by-step explanation:

Solution:-

- The random variable "X" denotes:

X : life expectancies of a certain protozoan

- The variable "X" follows normal distribution.

X ~ Norm (48, 10.5^2)

- A sample of n = 36 days was taken.

- The sample is also modeled to be normally distributed:

x ~ Norm (48, (10.5 / √n) ^2)

- The sample standard deviation s = 10.5 / √n = 10.5 / √36

s = 1.75

- We are to investigate the the probability of sample mean x_bar ≥ 51 days:

P (x_bar ≥ 51)

- Standardize the results, evaluate Z-score:

P (Z ≥ (x_bar - u) / s) = P (Z ≥ (51 - 48) / 1.75)

P (Z ≥ 1.7142).

- Use the standardized normal table and evaluate:

P (Z ≥ 1.7142) = 0.0432

Hence, P (x_bar ≥ 51) = 0.0432