Normal Probability Distributionmean = npstandard deviation = sqrt(np(1-p))The unemployment rate is 5.8%. Suppose that 100 employable people are selected randomly.a. What is the expected number of people who are unemployed?b. What are the variance and standard deviation of the number of people who are unemplyed?c. What is the probability that exactly six people are unemployed?d. What is the probability that at least four people are unemployed?

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Asked Nov 10, 2019
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Normal Probability Distribution

mean = np

standard deviation = sqrt(np(1-p))

The unemployment rate is 5.8%. Suppose that 100 employable people are selected randomly.

a. What is the expected number of people who are unemployed?

b. What are the variance and standard deviation of the number of people who are unemplyed?

c. What is the probability that exactly six people are unemployed?

d. What is the probability that at least four people are unemployed?

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Step 1

Hello! As you have posted 4 sub parts, we are answering the first 3 sub-parts.  In case you require the unanswered parts also, kindly re-post that parts separately.

Step 2

a.

 

From the given information, there are 100 employable people  and 5.8% of unemployment rate.

Let us define the random variable X as the number of people who are unemployed and follows Binomial distribution with parameters n = 100 and p = 0.058(=5.8%).

 

Formula for mean of binomial distribution is µ=np=100*0.058=5.8, approximately equal to 6.

 

Thus, the expected number of people who are unemployed is 6.

 

b.

 

The variance of the number of people who are unemployed is 5.4636 and it is calculated below:

Variance np (1- p)
100x 0.058x(1-0.058)
=5.4636
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Variance np (1- p) 100x 0.058x(1-0.058) =5.4636

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Step 3

The standard deviation of the number of people who are unemployed i...

Standard deviati on = ,fnp(1-p)
100 x0.058x (1-0.058)
= 5.4636
2.3374
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Standard deviati on = ,fnp(1-p) 100 x0.058x (1-0.058) = 5.4636 2.3374

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