1. Normal Distribution

Hope some one can help me with this

Let X1,X2 .... X16 be a random sample from a normal distribution N(77,25).

Compute a)P(77<X<79.5)
b) P(74.2<X<78.4)
Cheers

2. Let X1,X2 .... X16 be a random sample from a normal distribution N(77,25).

Compute a)P(77<X<79.5)
b) P(74.2<X<78.4)
What is X? the sample mean?

RonL

3. Yes -- it's actually X with a little line over the top- i didnt know how to put it in here.

Cheers

4. Let X1,X2 .... X16 be a random sample from a normal distribution N(77,25).

Compute a)P(77<X<79.5)
b) P(74.2<X<78.4)
The sample mean of a sample of size $\displaystyle n$ from a normal distribution
$\displaystyle N(\mu,\sigma^2)$ has a normal distribution; $\displaystyle N(\mu,\sigma^2/n)$.

So in this case the SD of the sample mean is $\displaystyle 5/4$ so:

$\displaystyle P(77<X<79.5)=P(\bar X<X<\bar X + 2.5)=$$\displaystyle P(\bar X<X<\bar X + 2 \sigma) , Which may be looked up in a suitable table: \displaystyle P(77<X<79.5)=P(\bar X<X<\bar X + 2 \sigma)=0.4772 . The second question is: \displaystyle P(74.2<X<78.4)=P(\bar X-2.8<X<\bar X + 1.4)=$$\displaystyle P(\bar X-2.24 \sigma <X<\bar X + 1.12 \sigma)$

Which from tables I make $\displaystyle 0.8686-0.0125=0.8561$

RonL