Question

Proctor & Gamble reported that an American family of four washes an average of 1 ton...

Proctor & Gamble reported that an American family of four washes an average of 1 ton (2000) pounds of clothes each year. If the standard deviation of the distribution is 187.5 pounds, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1980 and 1995 pounds. Round final answer to 4 decimal places (0.0001).

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Answer #1

Solution :

Given that ,

mean = = 2000

standard deviation = = 187.5

= / n = 187.5 / 50 = 26.5165

= P[(1980 - 2000) /26.5165 < ( - ) / < (1995 - 2000) / 26.5165)]

= P(-0.75 < Z < -0.19)

= P(Z < -0.19) - P(Z < -0.75)

= 0.4247 - 0.2266

= 0.1981

Probability = 0.1981

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