I was thinking P (Z < -1.61) = 1 - P( Z < 1.61) = xxx Does that make sense?
yes that will work
depends on table you use if table has neg Z values, then just look up Z=-1.61
What if P(Z < - 51)?
that is essentially 0....about 99% of distribution is within 3 std dev
I also got 0, but couldn't believe it :( Thanks!!
yw
Just to make sure I was working correctly: \[\mu = 40\]\[\sigma^2=38.4\]\[\sqrt{n}=\sqrt{1000}\] Find P(X<30) \[Z = \frac{30-40}{\sigma/\sqrt{n}}=-51\]P(X<30) ~= P (Z<-51) = 1-P(51.03) = 1-1 =0?
@dumbcow Sorry for interrupting :(
no prob didn't see you posted
Could you please check to see if my work makes sense?
is the variance a "sample variance" ?
I found the variance by \[\sigma^2 = npq\]
ok seems like it all checks out :) due to large sample size, the Z value is very high
Is 1000 a large sample size?
yes , usually anything over 50 is considered large or "significant"
Wow! Thanks a ton!!! :)
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