Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7 and standard deviation 0.75. (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.7 and 3.00?b) How large a sample size would be required to ensure thatthe first probability in part (a) is at least .99 ??

Answer 1

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable X ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.7 and standard deviation, σ = 0.75.

(a)

A random sample of n = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

$P(\bar X\leq 3.00)=P(\bar X$

                    $=P(\bar X$

                    $=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

                    $=P(Z$

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

$P(2.70$

                                $=P(0$

*Use a z-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

$P(\bar X\leq 3.00)\geq 0.99$

$P(\bar X$

$P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$P(Z$

The value of z for the probability above is

z ≥ 2.33

Compute the value of n as follows:

$\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33$

$\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}$

$\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77$

Thus, the sample size must be at least 77.