Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92.
A. If a random sample of 25 specimens is selected, what is theprobability that the sample average sediment density is at most3.00? Between 2.68 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:

The sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92.

Let $\bar X$ = sample  average sediment density

The z-score probability distribution for the sample mean is given by;

                             Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean = 2.68

           $\sigma$ = population standard deviation = 0.92

           n = sample of specimens = 25

(a) The probability that the sample average sediment density is at most 3.00 is given by = P($\bar X$ $\leq$ 3.00)

   

   P($\bar X$ $\leq$ 3.00) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z $\leq$ 1.74) = 0.9591

The above probability is calculated by looking at the value of x = 1.74 in the z table which has an area of 0.9591.

Also, the probability that the sample average sediment density is between 2.68 and 3.00 is given by = P(2.68 < $\bar X$ < 3.00)

P(2.68 < $\bar X$ < 3.00) = P($\bar X$ < 3.00) - P($\bar X$ $\leq$ 2.68)

 

   P($\bar X$ < 3.00) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z < 1.74) = 0.9591

   P($\bar X$ $\leq$ 2.68) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.68-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.74 and x = 0 in the z table which has an area of 0.9591 and 0.50.

Therefore, P(2.68 < $\bar X$ < 3.00) = 0.9591 - 0.50 = 0.4591.

(b) Now, we have to find a sample size that would ensure that the first probability in part (a) is at least 0.99, that is;

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

      P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }$ ) $\geq$ 0.99

      P(Z $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }$ ) $\geq$ 0.99

Now, in the z table; the critical value of x which has an area of at least 0.99 is given by 2.3263, that is;

                 $\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }=2.3263$

                 $\sqrt{n} } }=\frac{ 2.3263\times 0.92}{0.32}$

                  $\sqrt{n} } }=6.69$

                    n = 44.76 ≈ 45          {By squaring both sides}

Hence, the sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.