Question

Let the mean salary of workers be $920 and the standard deviation of salaries be $870....

Let the mean salary of workers be $920 and the standard deviation of salaries be $870. The probability that the average salary of the 100 workers is larger than $1000 is 17.88%.

Ture or False?

Math   Statistics-And-Probability   
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Answer #1

Solution :

= / n = 870 / 100 = 87

P( > 1000) = 1 - P( < 1000)

= 1 - P[( - ) / < (1000 - 920) / 87]

= 1 - P(z < 0.92)

= 0.1788

= 17.88%

True

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