Find the probability and interpret the results If convenient

Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was $2.709 per gallon. A random sample of 35 gas stations is drawn from this population. What is the probability that the mean price for the sample was between $2.699 and $2.724 that week? Assume sigma = $0.043. The probability that the sample mean was between $2.699 and $2.724 is . A. About 10% of the population of 35 gas stations that week will have a mean price between $2.699 and $2.724. B. About 90% of the sample of 35 gas stations that week will have a mean price between $2.699 and $2.724. C. About 10% of the sample of 35 gas stations that week will have a mean price between $2.699 and $2.724. D. About 90% of the population of 35 gas stations that week will have a mean price between $2.699 and $2.724.

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2.699      
x2 = upper bound =    2.724      
u = mean =    2.709      
n = sample size =    35      
s = standard deviation =    0.043      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.375832508      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.063748762      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.084436747      
P(z < z2) =    0.980479228      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.8960   [ANSWER]

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b)

OPTION B: About 90% of samples of 35 gas stations that week will have a mean pricebetween 2.699 and 2.724. [ANSWER, B]