# 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?

Question
46 views

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?

check_circle

star
star
star
star
star
1 Rating
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:

Step 3

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

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

### Other 