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The average starting salary for this year's graduates at a large university (LU) is $ 25...

The average starting salary for this year's graduates at a large university (LU) is $ 25 thousand with a standard deviation of $8 thousand. Furthermore, it is known that the starting salaries are normally distributed. What is the probability that a randomly selected LU graduate will have a starting salary of at least $ 21 thousand?

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Answer #1

Solution :

Given ,

mean = = 25

standard deviation = = 8

P(x > 21) = 1 - P(x< 21)

= 1 - P[(x -) / < (21-25) /8 ]

= 1 - P(z < -0.5)

Using z table

= 1 - 0.3085

= 0.6915

probability= 0.6915

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