Question

Suppose the sediment density(g/cm) of a randomly selected specimenfrom a certain region is normally distributed with mean 2.65 andstandard deviation .85.

a) If a random sample of 25 specimens is selected, what is theprobability that the sample average sediment density is at most3.00? Between 2.65 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

ai )
`P(\= X \le 3.0 ) =0.980`

aii)
`P(2.65 \le \= X \le 3.00) = 0.480`

b )
`n = 32`

Explanation:

From the question we are told that

The mean is
`\mu = 2.65`

The standard deviation is
`\sigma = 0.85`

Let the random sediment density be X

given that the sediment density is normally distributed it implies that

X N(2.65 , 0.85)

Now probability that the sample average is at 3.0 is mathematically represented as

`P(\= X \le 3.0 ) = P[\frac{\= X - \mu} {(\sigma )/(√(n) ) } \le (3.0 - \mu)/(( \sigma)/(√(n) ) ) ]`

Here n is the sample size = 25 and
`\= X` is the sample mean

Now Generally the Z-value is obtained using this formula

`Z = \frac{\= X - \mu} {(\sigma )/(√(n) ) }`

Thus

`P(\= X \le 3.0 ) = P[Z \le (3.0 - 2.35)/(( 0.85)/(√(25) ) ) ]`

`P(\= X \le 3.0 ) = P[Z \le 2.06 ]`

From the z-table the z-score is 0.980

Thus

`P(\= X \le 3.0 ) =0.980`

Now probability that the sample average is between 2.65 and 3.00 is mathematically evaluated as

`P(2.65 \le \= X \le 3.00) = P [(2.65 - \mu )/( (\sigma )/(√(n) ) ) < (\= X - \mu )/( (\sigma )/(√(n) ) ) < (3.0 - \mu )/( (\sigma )/(√(n) ) )]`

`P(2.65 \le \= X \le 3.00) = P [(2.65 - 2.65 )/( (0.85 )/(√(25) ) ) <Z < (3.0 - 2.65 )/( (0.85 )/(√(25) ) )]`

`P(2.65 \le \= X \le 3.00) = P[0 < Z< 2.06]`

`P(2.65 \le \= X \le 3.00) = P(Z < 2.06) - P(Z<0)`

From the z-table

`P(2.65 \le \= X \le 3.00) = 0.980 - 0.50`

`P(2.65 \le \= X \le 3.00) = 0.480`

Now from the question

`P(\= X \le 3.0 ) =0.99`

=>
`P(\= X \le 3.0 ) = P[Z \le (3.0 - 2.35)/(( 0.85)/(√(n) ) ) ] = 0.99`

Generally the critical value of z for a one tail test such as the one we are treating that is under the area 0.99 is
`t_z = 2.33` this is obtained from the critical value table

So

`t_z = (3.0 - 2.35)/(( 0.85)/(√(n) ) )`

`2.33 = (3.0 - 2.35)/(( 0.85)/(√(n) ) )`

=>
`n = 32`