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erastova [34]
4 years ago
13

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

ndstandard 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 ??
Mathematics
1 answer:
antiseptic1488 [7]4 years ago
4 0

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

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