Answer:
0.8413 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.00 centimeters
Standard Deviation, σ = 0.60 centimeters
Sample size, n = 16
We are given that the distribution of average diameter is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.60}{\sqrt{16}} = 0.15[/tex]
P(diameter of sample is more than 3.85 centimeter)
P(x > 3.85)
[tex]P( x > 3.85) = P( z > \displaystyle\frac{3.85 - 4}{0.15}) = P(z > -1)[/tex]
[tex]= 1 - P(z \leq -1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 3.85) = 1 -0.1587 = 0.8413[/tex]
0.8413 is the probability that the the average diameter of sample of 16 sand dollars is more than 3.85 centimeters.
