Answer:
0.918 is the probability that the sample average sediment density is at most 3.00
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 2.80
Standard Deviation, σ = 0.85
Sample size,n = 35
We are given that the distribution of sediment density 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{0.85}{\sqrt{35}} = 0.1437[/tex]
P(sample average sediment density is at most 3.00)
[tex]P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \leq 3.00) = 0.918[/tex]
0.918 is the probability that the sample average sediment density is at most 3.00
