Question

These data can be approximated quite well by a N(3.4, 3.1) model. Economists become alarmed when productivity decreases. According to the normal model what 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?

Answer 1

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:

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

We have to evaluate:

$P(x > \mu-0.5(\sigma)\\P(x > 3.4-0.5(3.1))\\P(x> 1.85)$

Evaluation of probability:

$P( x > 1.85) = P( z > \displaystyle\frac{1.85 - 3.4}{3.1}) = P(z >-0.5)$

$= 1 - P(z \leq -0.5)$

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

$P(x > 1.85) = 1 - 0.3085=0.6915 = 69.15\%$

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.