3. 0/3 points | Previous Answers DevoreStat9 5E.054. Suppose the sediment densit

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3. 0/3 points | Previous Answers DevoreStat9 5E.054. Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.6 and standard deviation 0.92. (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 2.6 and 3.00? (Round your answers to four decimal places.) at most 3.00 between 2.6 and 3.00 0,4772 0.9772 (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) 0.9777 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

= 2.6 = 0.92

(a) n = 25

x' = + z / n

x' = 3

=> 3 = 2.6 + 0.92z / 5

=> z = (3 - 2.6) * 5 / 0.92 = 2.1739.

P(z <= 2.1739) from tables = 0.9851.

P(2.6 <= z <= 3) = 0.9851 - 0.5

= 0.4851.

(b) A probability of 0.99 corresponds to z value 2.325.

=> 3 = 2.6 + 0.92*2.325 / n

=> n = 0.92*2.325 / (3 - 2.6) = 5.3475

n = 5.3475 * 5.3475 = 28.5958.

Thus the sample size must be 29.

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