Question

5.According to the Bureau of Labor Statistics, in calendar year 2000, the average duration of unemployment was 12.7 weeks, with a standard deviation of 0.3 weeks. Assuming that distribution is approximately normal, what is the probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks?

Answer 1

Answer:

Probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is 0.83144 .

Step-by-step explanation:

We are given that according to the Bureau of Labor Statistics, in calendar year 2000, the average duration of unemployment was 12.7 weeks, with a standard deviation of 0.3 weeks.

Let X = randomly selected individual who was unemployed in 2000

Since distribution is approximately normal, so X ~ N($\mu=12.7,\sigma^{2} = 0.3^{2}$)

The z score probability distribution is given by;

                 Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 12.7 weeks

            $\sigma$ = population standard deviation = 0.3 weeks

So, the probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is given by = P(12 < X < 13) = P(X < 13) - P(X <= 12)

P(X < 13) = P( $\frac{X-\mu}{\sigma}$ < $\frac{13-12.7}{0.3}$ ) = P(Z < 1) = 0.84134

P(X <= 12) = P( $\frac{X-\mu}{\sigma}$ < $\frac{12-12.7}{0.3}$ ) = P(Z < -2.33) = 1 - P(Z <= 2.33) = 1 - 0.99010

                                                                                                   = 0.0099

Therefore, P(12 < X < 13) = 0.84134 - 0.0099 = 0.83144 .