Respuesta :
Answer:
The probability that the sample average sediment density is at most 3.00 is 0.2090.
Step-by-step explanation:
Let X = he sediment density (g/cm) of a specimen from a certain region.
The random variable X is normally distributed with mean, μ = 2.63 and standard deviation, σ = 0.80.
A random sample of n = 25 specimens is selected.
The distribution of the sample mean of a normally distributed random variable is also normal.
To compute the probability of the sample mean we need to first compute the z score of the sample mean value. The formula of z-score is:
[tex]z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}[/tex]
Compute the probability that the sample average sediment density is at most 3.00 as follows:
Apply continuity correction:
[tex]P(\bar X\leq 3.00)=P(\bar X< 3.00-0.50)[/tex]
[tex]=P(\bar X<2.50)\\=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{2.50-2.63}{0.80/\sqrt{25}})[/tex]
[tex]=P(Z<-0.81)\\=1-P(Z<0.81)\\=1-0.79103\\=0.20897\\\approx0.2090[/tex]
*Use a z-table for the probability.
Thus, the probability that the sample average sediment density is at most 3.00 is 0.2090.
