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