Answer:
Option A) 0.0127
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 16 weeks
Standard Deviation, σ = 2 weeks
Sample size = 20
We are given that the distribution of time taken to find a job is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P(20 young workers average less than 15 weeks)
P(x < 15)
[tex]P( x < 15) = P( z < \displaystyle\frac{15-16}{\frac{2}{\sqrt{20}}}) = P(z < -2.236)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 15) =0.0127= 1.27\%[/tex]
Thus, 0.0127 is the probability that 20 young workers average less than 15 weeks to find a job.
