Hope some one can help me with this

Cheers ;)Quote:

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)

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- Apr 22nd 2006, 07:05 PMdarren_a1Normal Distribution
Hope some one can help me with this

Quote:

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)

- Apr 23rd 2006, 12:26 AMCaptainBlackQuote:

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)

RonL - Apr 23rd 2006, 06:24 AMdarren_a1
Yes -- it's actually X with a little line over the top- i didnt know how to put it in here.

Cheers - Apr 23rd 2006, 08:13 AMCaptainBlackQuote:

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)

$\displaystyle N(\mu,\sigma^2)$ 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