7. Probability computations using the standard normal distribution AaAa Assume t

Question

7. Probability computations using the standard normal distribution AaAa Assume that X, the starting salary offer for psychology majors, is normally distributed with a mean of $47,507 and a standard deviation of $5,000. Use the following Distributions tool to help you answer the questions. (Note: To begin, click on the button in the lower left hand corner of the tool that displays the distribution and a single orange line.) Standard Normal Distribution Mean o.0 Standard Deviation = 1.0 -4 -2 The probability that a randomly selected psychology major received a starting salary offer greater than $45,000 is The probability that a randomly selected psychology major received a starting salary offer between $45,000 and $52,000 is (Hint: The standard normal distribution is perfectly symmetrical about the mean, the area under the curve to the left (and right) of the mean is 0.5. Therefore, the area under the curve between the mean and a z-score is computed by subtracting the area to the left (or right) of the z-score from 0.5.)

Explanation / Answer

mean = 47507 , s = 5000

P(x > 45000)
z = ( x -mean) / s
= ( 45000 - 47507) / 5000
= -0.5014
P(x > 45000) = P(Z > -0.5014) = 0.692 By using standard table

P( 45000 < x < 52000)
= P[( x -mean) / s < z < ( x -mean) / s ]
= P[( 45000 - 47507) / 5000 < z < ( 52000 - 47507) / 5000]
= P(-0.5014 < z < 0.8986)
P( 45000< x < 52000) = P(-0.5014 < z < 0.8986) = 0.5075 By using standard table

P( 38000 < x < 45000)
= P[( x -mean) / s < z < ( x -mean) / s ]
= P[( 38000 - 47507) / 5000 < z < ( 45000 - 47507) / 5000]
= P(-1.9014 < z < -0.5014)
P( 38000< x < 45000) = P(-1.9014 < z < -0.5014) = 27.98% By using standard table

z value at 20% less than = -0.8416

z =(x -mean) / s
-0.8416 = ( x - 47507)/5000
x = 43299

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