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

Answer:

ai )  $P(\= X \le 3.0 ) =0.980$

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

b )   $n = 32$

Step-by-step 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

        $P(\= X \le 3.0 ) = P[\frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } } \le \frac{3.0 - \mu}{\frac{ \sigma}{\sqrt{n} } } ]$

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

  Now  Generally the  Z-value is obtained using this  formula  

           $Z = \frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } }$

Thus  

       $P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{25} } } ]$

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

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

           $P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{\= X - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{3.0 - \mu }{ \frac{\sigma }{\sqrt{n} } }]$  

          $P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - 2.65 }{ \frac{0.85 }{\sqrt{25} } }$

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

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

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  

         $P(\= X \le 3.0 ) =0.99$

=>       $P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } } ] = 0.99$

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  

        $t_z = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

        $2.33 = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

=>       $n = 32$