Question

Help with four elementary statistics questions..

Answer 1

A normal distribution is always centered about its mean, so you can eliminate C as an option.

If you're familiar with this rule, https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule then you might know that most of the distribution (a little over \(99\%\)) lies within \(3\) standard deviations from the mean. Since the standard deviation here is \(2\), this means \(99\%\) of the distribution should fall between \(20-3(2)=14\) and \(20+3(2)=26\).

Which option do you think satisfies the rule?

Answer 2

D? Im not familiar with any of that. I also dont trust Wikipedia as a source.

Answer 3

D is correct. If you're not familiar with that rule, have you learned about any other ways of identifying a normal distribution only knowing the mean and standard deviation? Roughly speaking, the mean tells you where the distribution is centered, while the standard deviation tells you how wide it is, but if B was a bit less wide it would be difficult to choose between it and D.

Also, no need to mistrust Wikipedia on something as well-established as this rule. It's a fairly common part of basic statistics curricula.

Answer 4

My course is online with no book. I am only given a handful of homework assignments and one practice quiz.

Answer 5

Hmm, well for future reference, I guess you can pick the right graph by thinking "large std dev = wide distribution" while "small std dev = narrow distribution".

For this next question, if \(X\) represents a woman's height, then you're looking for the probability that \(X>70\), which is where the \(z\) score comes into play.

You have to first the normal distribution for \(X\) for the standard normal distribution with mean \(0\) and std dev \(1\). You do this by writing
\[Z=\frac{X-\mu}{\sigma}\implies X=\mu+\sigma Z\]where \(\mu\) is the mean and \(\sigma\) is the std dev of \(X\). So you have
\[Z=\frac{X-65}{2.5}\implies X=65+2.5Z\]You want the probability that \(X>70\), or
\[X>70\implies65+2.5Z>70\implies Z>2\]Do you know how to find this probability with a \(z\) score of \(2\)?

Answer 6

Sorry I was writing down what you said about the previous question.. It would be 0.97725

Answer 7

Not quite. I'm guessing you have a table that uses left-endpoint probabilities, which actually give \(\mathbb P(Z<k)\), like the one here: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
You're interested in \(\mathbb P(Z>k)\), which you can find by either looking for a right-endpoint table or taking advantage of the fact that all probability distributions add up to \(1\). The second method is quicker, as this means
\[\mathbb P(Z<k)+\mathbb P(Z>k)=1\implies \mathbb P(Z>2)=1-\mathbb P(Z<2)\approx0.0228\]

Answer 8

Okay. Now I have 0.0228. Whats the next step?

Answer 9

There is no next step. That's the probability you're looking for in the second question (which you seem to have deleted). There's an approximately \(2\%\) chance that women are taller than the average man, if I remember correctly.