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.

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shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-04-25T05:19:23+02:00

Answer:

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