Question

In: Statistics and Probability

A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fare...

A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fare charged for service from Hermosa Beach to Los Angeles International Airport is $18 and the standard deviation is $3.00. We select a sample of 15 fares. What is the likelihood that the sample mean is between $17 and $19? Assume the population of fares is normally distributed. (Round z value to 2 decimal places and final answer to 4 decimal places.)

Solutions

Expert Solution

Solution :

Given that,

mean = = 18

standard deviation = = 3

n = 15

= = 18

= / n = 3 / 15 = 0.7746

P(17 < < 19) = P((17 - 18) / 0.7746 <( - ) / < (19 - 18) / 0.7746))

= P(-1.29 < Z < 1.29)

= P(Z < 1.29) - P(Z < -1.29) Using z table,

= 0.9015 - 0.0985

= 0.8030

orchestra answered 2 years ago
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