Suppose the sediment density (g/cm) of a randomly selected specimen from a certa

Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region Is normally distributed with mean 2.61 and standard deviation 0.74. If a random sample of 25 specimens Is selected, what is the probability that the sample average sediment density Is at most 3.00? Between 2.61 and 3.00? (Round your answers to four decimal places.) at most 3.00 between 2.61 and 3.00 How large a sample size would be required to ensure that the probability in part (a) is at least 0.99? (Round your answer up to the nearest whole number.)

Explanation / Answer

Mean ( u ) =2.61
Standard Deviation ( sd )=0.74
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 3) = (3-2.61)/0.74
= 0.39/0.74= 0.527
= P ( Z <0.527) From Standard Normal Table
= 0.7009                  
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 2.61) = (2.61-2.61)/0.74
= 0/0.74 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 3) = (3-2.61)/0.74
= 0.39/0.74 = 0.527
= P ( Z <0.527) From Standard Normal Table
= 0.70091
P(2.61 < X < 3) = 0.70091-0.5 = 0.2009  

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