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

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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437$

P(sample average sediment density is at most 3.00)

$P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)$

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