The number of years of education of self-employed individuals in the united states has a population mean of 13.6 years and a population standard deviation of 3.0 years. if we survey a random sample of 100 self-employed people to determine the average number of years of education for the sample, what is the standard deviation of the sampling distribution of x-bar (the sample mean)? 3.0 years 0.3 years 13.6 years

0.3 years

Step-by-step explanation:

With problems like these, I always like to start by breaking down the information into smaller pieces.

μ = 13.6

σ = 3.0

Survey of 100 self-employed people

(random variable) X = # of years of education

So now we have some notation, where μ represents population mean and σ represents population standard deviation. Hopefully, you already know that the sample mean of x-bar is the same as the population mean, so x-bar = 13.6. Now, the question asks us what the standard deviation is. Since the sample here is random, we can use the Central Limit Theorem, which allows us to guess that a distribution will be approximately normal for large sample sizes (that is, n ≥ 30). In this case, our sample size is 100, so that is satisfied. We're also told our sample is random, so we're good there, too. Now all we have to do is plug some stuff in.

The Central Limit Theorem says that for large values of n, x-bar follows an approximately normal distribution with sample mean = μ and sample standard deviation = σ/√n. So, with that info, all we need to do to find the standard deviation of x-bar is to plug our σ and n into the above formula.

σ(x-bar) = σ/√n

σ(x-bar) = 3.0/√100

σ(x-bar) = 0.3

So your answer here is .3 years.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: