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

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

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 =
`(\bar X-\mu)/((\sigma)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(3.00-2.68)/((0.92)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(3.00-2.68)/((0.92)/(√(25) ) )` ) = P(Z < 1.74) = 0.9591

P(
`\bar X`
`\leq` 2.68) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(2.68-2.68)/((0.92)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(3.00-2.68)/((0.92)/(√(n) ) )` )
`\geq` 0.99

P(Z
`\leq`
`(3.00-2.68)/((0.92)/(√(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;

`(3.00-2.68)/((0.92)/(√(n) ) )=2.3263`

`√(n) } }=( 2.3263* 0.92)/(0.32)`

`√(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.