Question

According to the Bureau of Labor Statistics it takes an average of 22 weeks for someone over 55 to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that eight workers over the age of 55 take an average of more than 20 weeks to find a job

Answer 1

Answer:

The probability that eight workers over the age of 55 will take an average of more than 20 weeks to find a job is 0.9977 or 99.77%

Step-by-step explanation:

Average time to find a new job for someone over 55 years = μ = 22 weeks

Standard deviation = σ = 2 weeks

We have to find the probability that if 8 workers are selected at random what will be the probability that it will take them more than 20 weeks to find a job. So, this means that the sample size is n = 8.

Since, the distribution is normal and we have the value of population standard deviation, we will use the z-distribution to find the desired probability. For this, first we need to convert the value (20 weeks) to its equivalent z-score. The formula to calculate the z-score is:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

x = 20, converted to z-score will be:

$z=\frac{20-22}{\frac{2}{\sqrt{8}}}=-2.83$

Thus, probability of time being greater than 20 weeks is equivalent to probability of z score being greater than - 2.83.

i.e.

P( X > 20 ) = P( z > -2.83 )

Using the z-table we can find this probability:

P( z > -2.83 ) = 1 - P( z < -2.83)

= 1 - 0.0023

= 0.9977

Since, P( X > 20 ) = P( z > -2.83 ), we can conclude that:

The probability that eight workers over the age of 55 will take an average of more than 20 weeks to find a job is 0.9977 or 99.77%