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 ??

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

Here n is the sample  size  = 25 and   is the sample  mean  

  Now  Generally the  Z-value is obtained using this  formula  

Thus  

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  

=>      

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  

=>      

answer: i think the answer is etheir c or d but i don't know which one but have a good day!

step-by-step explanation:

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!