The Crossett Trucking Company claims that the mean weight of their delivery truc

Question

The Crossett Trucking Company claims that the mean weight of their delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed. a. Within what limits will the middle 95 percent of the sample means occur?_______________________ 3. A recent survey of 50 employed male executives showed that it took an average of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. a. Construct a 95 percent confidence interval for the population mean. _____________ b. Is it reasonable that the population mean is 28 weeks? __________________ c. Justify your answer in b_____________________________________________________ d. State the margin of error and critical error._

Explanation / Answer

Q2.

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 6000
standard Deviation ( sd )= 150
LESS THAN
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 6000/150 ) = 0.025
That is, ( x - 6000/150 ) = -1.96
--> x = -1.96 * 150 + 6000 = 5706.0054
GREATER THAN
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u / (s.d) > x - 6000/150) = 0.025
That is, ( x - 6000/150) = 1.96
--> x = 1.96 * 150+6000 = 6293.9946

With in the middle 95 percent of the sample means occur is [ 5706.0054 < x < 6293.9946]

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