# Thread: Distribution and probabilty - parent and sample

1. ## Distribution and probabilty - parent and sample

Ok I am completely lost, can anyone help me out? I have done the first question, and the second one is based on it.

The first question said:
The parent distribution of x is normal with mean 80 and standard deviation 9.
Based on samples of size n=25, find the mean and standard error of the sampling distribution of $\bar{x}$.

I said: $\mu_{\bar{x}}$ = 80
$\sigma_{\bar{x}}$ = 9/5

The second question is where I am lost.

It says find:
A. P(62 < x < 80)
B. P(71 < $\bar{x}$ < 77)
C. x' such that P( $\bar{x}$ > x') = .05

2. Is n=25 or 225? Because $\sigma_{\bar{x}}={\sigma \over\sqrt{n}}$.

Or, did you mean $\sigma_{\bar{x}}={9\over 5}$

3. n=25

it is the number in the sample
$\sigma\bar{x}$ = 9/5 (thats what I got) sorry

4. $P(62.

$P(71<\bar X<77)=P\biggl({71-80\over 9/5}.

Can you finish these? I'll check on you later.

$.05=P(\bar X so ${a-80\over 9/5}=-1.645$.

5. P[(-5) < z < (-5/3)]

It was supposed to be P( $\bar{x}$ > x') SORRY

P( $\bar{x}$ > $x'-80\over 9/5$) ?

Can you tell me what formula you used to find that last one?

Did you use these for the first two?
Z= $X-\mu\over \sigma$ and Z= $\bar{x}-\mu\over \sigma / \sqrt n$

6. Let X be a random variable with mean $\mu$ and standard deviation $\sigma$.
Then $Z={X-\mu\over \sigma}$ has mean zero and standard deviation 1.
And if X was a normal random variable, then Z is a standard normal random variable and you can use your Z table.
We did that for both X and $\bar X$.

7. Thank you SO much for your help. It helped me a lot on my test. Thanks a hundred times. You are a really nice person