Question

The mean per capita income is 18,904 dollars per annum with a standard deviation of 469...

The mean per capita income is 18,904 dollars per annum with a standard deviation of 469 dollars per annum. What is the probability that the sample mean would be less than 18850 dollars if a sample of 333 persons is randomly selected? Round your answer to four decimal places.

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Answer #1

Solution :

Given that ,

mean = \mu = 18904

standard deviation = \sigma = 469

\sigma\bar x = \sigma / \sqrt n = 469 / \sqrt 333 = 25.7010

P(\bar x < ) = P((\bar x - \mu \bar x ) / \sigma \bar x < (18850 - 18904) / 25.7010)

= P(z < -2.1011)

= 0.0178

probability = 0.0178

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