Help with Statistics! due today!

According to a report, 56.2% of murders are committed with a firearm.
(a) if 200 murders are randomly selected, how many would we expect to be committed with a firearm? = 112
(b) Would it be unusual to observe 134 murders by firearm in a random sample of 200 murders? Why?

Help me! it's my last question and i need help.


For (b)-
No, because 134 is between u-2o and u+2o
No, because 134 is greater than u+2o
Yes, because 134 is greater than u+2o
Yes, because 134 is between u-2o and u+2o
No, because 134 is less then u-2o

Your disribution is Binomial with mean 200(.562)=112.4
and Variance112.4(1-.562)=49.2312
so the standard deviation is sqrt(Variance)=7.0165
So the mean + 2 standard deviations
=112.4+2(7.0165)=126.43
so your answer is yes because 134 is greater than u+2o

μ = 56.2%
(a). Expected value from random sample of 200 = 200 x 0.562 = 112 rounded
(b). You cannot answer this without knowing σ

134 -112 = 22. If 22 > 2σ ( σ < 11). only about 35 of values fall in that range, so you would consider it unusual., so it's "Yes, because 134 is greater than u+2o"

See the handy graphic at the start of:

http://en.wikipedia.org/wiki/File:Standa…

what is 'u' ?
If its variance, 75% of data will lie in between
Mean + - variance

112 +- variance
Or U CAN USE C.I. Confidant intervel