5.According to the Bureau of Labor Statistics, in calendar year 2000, the average duration of unemployment was 12.7 weeks, with a standard d

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?

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Allison 2 months 2021-10-16T06:54:27+00:00 1 Answer 0 views 0

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    2021-10-16T06:55:43+00:00

    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 .

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