Question

Lets say that ages in this class are distributed normally with a mean of 19 and a standard deviation of 2, what does this mean? Using this data what is the likelihood that any randomly selected group of five students has an average age of 20 or older. With the same data lets now assume that the class is a random sample of all University level classes and that there are 50 people in the class. What is the 95% confidence interval for your estimation of ages for university level students? Show any of your work but answer this as if you are teaching it to an 8th grade class.

Answer 1

As the ages are distributed normally we can convert x into z.

Here we need to find $P(x\geq 20)$

$P\left ( z\geq \frac{20-19}{2} \right )=P(z\geq 0.5)=0.309$

ん

So the likelihood that any randomly selected group of five students has an average age of 20 or older are 5*0.309=1.545

Here we need to find 95% confidence interval with n=50

Now z score for 95% confidence interval is 1.96 as P(-1.96<z<1.96)=0.95

Margin of error $E=z*\frac{\sigma}{\sqrt{n}}$ $E=z*\frac{\sigma}{\sqrt{n}}=1.96*\frac{2}{\sqrt{50}}=0.554$

95%Confidence interval $CI=\mu\pmE=19\pm0.554=(18.446,19.554)$