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The average hourly wage of employees of a certain company is $12.96. Assume the variable is normally distributed. If the standard deviation is $4.46, find the probability, correct to the nearest thousandth, that a randomly selected employee earns more than $15.00 per hour.
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we know mean =12.96 st D =4.46 so we are finding P(x > 15) alright so we are going to apply the central limit theorem right here so we got p((x - mean)/st. dev > (15 - 12.96)/4.46 which is now P(Z > 0.46) which is approximately 0.3227 :)
thanks ;))
lol no problem
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