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 2.60 and standard deviation 0.88. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? (Give answer accurate to 3 decimal places.) What value of x for sediment density is such that the sample mean is at most x with probability .99? (Give answer accurate to 2 decimal places.) How large a sample size would be required to ensure that the probability in part (a) is at least .99? n = 33.9

Explanation / Answer

a) here sample mean =2.6

and std error of mean =std deviation/(n)1/2 =0.176

hence P(X<3)=P(Z<(3-2.6)/0.176)=P(Z<2.2727)=0.988

b)for 0.99 . z=2.326

hence x=mean +z*Std error =3.009

c) for that std error =0.4/2.326 =0.1719

hence sample size =(0.88/0.1719)2 =27

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