3. Given f(x)=-2(5) Find f (4)
1 answer:
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Answer:
Step-by-step explanation:
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()
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 .