Question 133641
The average afe of CEO's is 56. Assume the cariable is normally distributed. If the standard deviation is 4 years, find the probability that the age of a randomly selected CEO will be in the following range:
Between 53 and 59 years old. 
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Find the z-score for 53 and for 59:
z(53) = (53-56)/4 = -0.75
z(59) = (59-56)/4 = 3/3 = +0.75
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P(53 < x < 59) = P(-.075 < z < 0.75) = 0.5467
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Let me know if you do not understand all of this.
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If you have a TI calculator use normalcdf(53,59,56,4) = 0.5467
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Cheers,
Stan H.