Answer :
Given:
μ = 5.3 cm, population mean
σ = 1.8 cm, population sandard deviation
n = 5, sample size
The random variable is x = .5 cm.
The z-score is
[tex]z = \frac{x-\mu}{\sigma / \sqrt{n} } = \frac{5.5-5.3}{1.8/ \sqrt{5} } =0.2485[/tex]
From standard table, obtain
P(x>5.5) = 1 - 0.598 = 0.402
Answer: 0.402
μ = 5.3 cm, population mean
σ = 1.8 cm, population sandard deviation
n = 5, sample size
The random variable is x = .5 cm.
The z-score is
[tex]z = \frac{x-\mu}{\sigma / \sqrt{n} } = \frac{5.5-5.3}{1.8/ \sqrt{5} } =0.2485[/tex]
From standard table, obtain
P(x>5.5) = 1 - 0.598 = 0.402
Answer: 0.402
Answer:
Step-by-step explanation:
Let X be the basal diameter of a sea anemone.
Given that X is N(5.3, 1.8)
Sample size =5
We have to find the prob that all five anemones have a basal diameter more than 5.5 cm
Prob (for one anemone >5.5 cm) = [tex]P(Z>\frac{5.5-5.3}{1.8/\sqrt{5} } =0.2485[/tex]
Since each anemone selected is independent of the other anemone, the selected anemones follow Binomial with n =5 and p = 0.2485 and q=0.7515
P(X=5) = 1-P(X=0) = 1-0.7515^5
=0.7603
Note that 5 elements selected cannot be taken as normal since sample size is small.
No of anemones selected having a basal diameter more than 5.5 cm is
binomial with n=5 and p = 0.2585