SOLUTION: Let x be a random variable that represents the amount of a randomly selected monthly cell phone bill. We know that x has a normal distribution with a mean of µ = $67.50 and a stan

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Question 889628: Let x be a random variable that represents the amount of a randomly selected monthly cell phone bill. We know that x has a normal distribution with a mean of µ = $67.50 and a standard deviation of σ = $16.09. What is the probability that the mean amount of five cell phone bills taken at random is between $63.50 and $71.50?
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Since the sample size is < 30, we have to calculate a std dev for the sample size of 5,
sample std dev = st dev of population / square root(5) = 16.09 / 2.236067977 = 7.195666753
now calculate z-values
z1-value = (63.50 - 67.50) / 7.195666753 = −0.555890112 = approx -0.56
z2-value = (71.50 - 67.50) / 7.195666753 = 0.555890112 = approx 0.56
P( 63.50 < X < 71.50 ) = P( X < 71.50 ) - P( X < 63.50)
consult z-table for associated probabilities
P( 63.50 < X < 71.50 ) = 0.7123 - 0.2877 = 0.4246