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 2.66 and standard deviation 0.73. (a) 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 2.66 and 3.00? (Round your answers to four decimal places.) at most 3.00 2.32 between 2.66 and 3.00 9843 (b) 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.) 2.73 X specimens You may need to use the appropriate table in the Appendix of Tables to answer this question Need Help? Read It Talk to a Tutor

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 2.66
standard Deviation ( sd )= 0.73/ Sqrt ( 25 ) =0.146
sample size (n) = 25
a.
P(X < 3) = (3-2.66)/0.73/ Sqrt ( 25 )
= 0.34/0.146= 2.3288
= P ( Z <2.3288) From Standard NOrmal Table
= 0.99006
P(X > = 3) = 1 - P(X < 3)
= 1 - 0.99006 = 0.00994
b.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 2.66) = (2.66-2.66)/0.73/ Sqrt ( 25 )
= 0/0.146
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 3) = (3-2.66)/0.73/ Sqrt ( 25 )
= 0.34/0.146 = 2.32877
= P ( Z <2.32877) From Standard Normal Table
= 0.99006
P(2.66 < X < 3) = 0.99006-0.5 = 0.49006
c.
P ( Z > x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is 2.326348
P( x-u / (s.d) > x - 2.66/0.146) = 0.01
That is, ( x - 2.66/0.146) = 2.326348
--> x = 2.326348 * 0.146+2.66 = 2.999647

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