Question

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50. Assume the distribution of hourly wages follows the normal probability distribution. We select a crew member at random.
What is the probability the crew member earns:


(a) Between $20.50 and $24.00 per hour (Round your answer to 4 decimal places.)


Probability


(b) More than $24.00 per hour (Round your answer to 4 decimal places.)


Probability


(c)
Less than $19.00 per hour (Round intermediate calculations to 2 decimal place. Round final answer to 4 decimal place)



Probability

Answer 1

Answer:

a) 0.3413

b) 0.1587

c) 0.3341

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 20.50 dollars

Standard Deviation, σ = 3.50 dollars

We are given that the distribution of hourly wages is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(Between $20.50 and $24.00 per hour

$P(20.50 \leq x \leq 24) = P(\displaystyle\frac{20.50 - 20.50}{3.50} \leq z \leq \displaystyle\frac{24-20.50}{3.50}) = P(0 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < 0)\\= 0.8413 - 0.5000 = 0.3413 = 34.13\%$

$P(20.50 \leq x \leq 24) = 34.1\%$

b) P(More than $24.00 per hour)

P(x > 24)

$P( x > 24) = P( z > \displaystyle\frac{24 - 20.50}{3.50}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,  

$P(x > 610) = 1 - 0.8413 = 0.1587 = 15.87\%$

c) P(Less than $19.00 per hour)

P(x < 19)

$P( x < 19) = P( z > \displaystyle\frac{19 - 20.50}{3.50}) = P(z < -0.4285)$

Calculation the value from standard normal z table, we have,  

$P(x < 19) =0.3341= 33.41\%$