Question

Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and standard deviation of 0.85. If a random sample of 35 specimens is selected, what is the probability that the sample average sediment density is at most 3.00?

Answer 1

0.918 is the probability that the sample average sediment density is at most 3.00

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:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling:

`=(\sigma)/(√(n)) = (0.85)/(√(35)) = 0.1437`

P(sample average sediment density is at most 3.00)

`P( x \leq 3.00) = P( z \leq \displaystyle(3.00 - 2.80)/(0.1437)) = P(z \leq 1.3917)`

Calculation the value from standard normal z table, we have,

`P(x \leq 3.00) = 0.918`

0.918 is the probability that the sample average sediment density is at most 3.00