Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.88. 

(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 2.65 and 3.00? 

(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.) 

Answer 1

given that mean = 2.65 , standard deviation = 0.88
(a)p(x ≤ 3.00) = p(z < (3.00 - 2.65)/(0.88/sqrt(25))
= p(z < 1.9886)
= 0.9767

p(2.65 < x < 3.00) = p(((2.65 - 2.65)/(0.88/sqrt(25)) < z < ((3.00 - 2.65)/(0.88/sqrt(25)))
= p(0 < z < 1.9886)
= p(z < 1.9886)− p(z < 0)
= 0.9767 - 0.5
= 0.4767

(b)given that value is atleast 0.99%
then z is 2.33
formula
=> z = (x - µ)/(σ / sqrt(n))
=> 2.33 = (3.00 - 2.65)/(0.88/sqrt(n))
=> 2.33 = 0.35/(0.88/sqrt(n))
=> 6.6571 = sqrt(n)/0.88
=> n = 34