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
