Answer:
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.
-3x + 6 = 18...subtract 6 from both sides
-3x = 18 - 6
-3x = 12...divide both sides by -3
x = -12/3 reduces to -4 <===
Qr=1/2 mn = ms
mn=2 ms=12
ln^2=lm^2+mn^2 using phythagoras rule
ln^2=400
ln=20
