Hope some one can help me with this
CheersLet 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 distributionLet 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)
$\displaystyle N(\mu,\sigma^2)$ 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