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 2.6 and standard deviation 0.83. (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 (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.) specimens

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3      
u = mean =    2.6      
n = sample size =    25      
s = standard deviation =    0.83      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.409638554      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.409638554   ) =    0.992015834 [ANSWER]

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We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2.6      
x2 = upper bound =    3      
u = mean =    2.6      
n = sample size =    25      
s = standard deviation =    0.83      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.409638554      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.992015834      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.492015834   [ANSWER]

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b)

For a 0.99 left tailed area, by table/technology,

z = 2.326347874

Hence,

n = [z*sigma/(x-u)]^2 = (2.326347874*0.83/(3.00-2.6))^2 = 23.30158796 = 24 [ANSWER]

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