Question

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Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,300 pounds and the standard deviation is 170 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed. Within what limits will 99% of the sample means occur? (Round your z-value to 2 decimal places and final answers to 1 decimal place.)

Answer 1

sd - 170 pounds

mean weight = 5300 pounds

sample N = 40 trucks

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step 1 : standard deviation of mean is:

s = σ/√n = 170/√40= 26.89

step 2:

99% of the sample means will be between 5300 -[ z(0.995) * 26.89] and 5300 + [z(0.995) * 26.89]

|z(0.995)| = 2.58

5300- (2.58*26.89) and 5300+ (2.58*26.89)

5230.6 and 5369.4 pounds are the limits of sample mean

FORMULA

Z = sample mean weight - mean weight

                      sd of mean

2.58 = sample mean weight - 5300  

                   26.89

2.58*26.89+5300 = SAMPLE MEAN WEIGHT = 5369.4 POUNDS .ROUNDING OFF

Z VALUE FOR 99% CONFIDENCE LEVEL IS 2.576 ROUNDING OFF TO 2.58