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?

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()

The z score probability distribution is given by;

                 Z = ~ N(0,1)

where, = population mean = 12.7 weeks

            = 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( < ) = P(Z < 1) = 0.84134

P(X <= 12) = P( < ) = 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 .

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3