Question

6. Find the area under the standard normal distribution curve. a. To the right of Z= .42 b. To the left of Z= -.5 c. Between Z= -.3 and Z= 1.2 8. The average annual salary of all US registered nurses in 2012 is $65,000. Assume the distribution is normal and the standard deviation is $4000. Find the probability that a random selected registered nurse earns.

a. Greater than $75,000

b. Between $50,000 and $70,000.

Answer 1

6)

This is a normal distribution question with

a)

z = 0.42

This implies that

P(z > 0.42) = 0.3372

b) z = -0.5

This implies that

P(z < -0.5) = 0.3085

c)

z1 = -0.3, z2 = 1.2

This implies that

P(-0.3 < z < 1.2) = 0.5028

8)

This is a normal distribution question with

a)

P(x > 75000.0)=?

The z-score at x = 75000.0 is,

z = 2.5

This implies that

P(x > 75000.0) = P(z > 2.5) = 1 - 0.9937903346742238

b)

P(50000.0 < x < 70000.0)=?

This implies that

P(50000.0 < x < 70000.0) = P(-3.75 < z < 1.25) = P(Z < 1.25) - P(Z < -3.75)

P(50000.0 < x < 70000.0) = 0.8943502263331446 - 8.841728520080377e-05

PS: you have to refer z score table to find the final probabilities.

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