Answer:
0.6915 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
N(3.4, 3.1)
Mean, μ = 3.4
Standard Deviation, σ = 3.1
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to evaluate:
[tex]P(x > \mu-0.5(\sigma)\\P(x > 3.4-0.5(3.1))\\P(x> 1.85)[/tex]
Evaluation of probability:
[tex]P( x > 1.85) = P( z > \displaystyle\frac{1.85 - 3.4}{3.1}) = P(z >-0.5)[/tex]
[tex]= 1 - P(z \leq -0.5)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 1.85) = 1 - 0.3085=0.6915 = 69.15\%[/tex]
0.6915 is the probability that the percent change in worker output per hour from the previous quarter is more than 0.5 standard deviations below the mean.
