Answer:
ai ) 
aii)
b ) 
Step-by-step explanation:
From the question we are told that
The mean is 
The standard deviation is 
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} {\frac{\sigma }{\sqrt{n} } } \le \frac{3.0 - \mu}{\frac{ \sigma}{\sqrt{n} } } ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%5Cle%203.0%20%29%20%3D%20P%5B%5Cfrac%7B%5C%3D%20X%20-%20%5Cmu%7D%20%7B%5Cfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%5Cle%20%5Cfrac%7B3.0%20-%20%5Cmu%7D%7B%5Cfrac%7B%20%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%20%5D)
Here n is the sample size = 25 and
is the sample mean
Now Generally the Z-value is obtained using this formula

Thus
![P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{25} } } ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%5Cle%203.0%20%29%20%3D%20P%5BZ%20%5Cle%20%5Cfrac%7B3.0%20-%202.35%7D%7B%5Cfrac%7B%200.85%7D%7B%5Csqrt%7B25%7D%20%7D%20%7D%20%20%5D)
![P(\= X \le 3.0 ) = P[Z \le 2.06 ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%5Cle%203.0%20%29%20%3D%20P%5BZ%20%5Cle%202.06%20%5D)
From the z-table the z-score is 0.980
Thus

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

From the z-table
Now from the question

=> ![P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } } ] = 0.99](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%5Cle%203.0%20%29%20%3D%20P%5BZ%20%5Cle%20%5Cfrac%7B3.0%20-%202.35%7D%7B%5Cfrac%7B%200.85%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%20%5D%20%3D%200.99)
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
this is obtained from the critical value table
So


=> 