check part C I got 0.8 the first time and the second time i got 1.3
Ps.- sorry i cut off part C a little bit, but it says 1.3
is your mean correct?
yes it should be
mean = (sum)/total numbers = (0+1+2+3+4)/5 isn't it ?
yes
mean = Just add up all the numbers, then divide by how many numbers there are
so isn't mean = 2 ?
i got this- 4*0.2 + 3*0.25 + 2*0.3 + 1*0.1 + 0*0.15
= 2.25
also thats not the way you calculate std deviation or mean thats just \(\sum x_i\times i\)
ok then im worng and your right
you have the formula of std deviation with u ?
Howard did it correctly when finding the mean when the values have different probabilities , you do what he did. and get 2.25 (the "simple formula" assumes equal probabilities = 1/n )
so B is 2.25 ?
mean "grade" was asked...thats why I thought
anyways, then yes, 2.25 is correct...sorry
i think it means grade level i hope but 0 isnt a grade lvl
ok, now std deviation N=5, (x_i)s are your P(X) values \(\mu \) is the mean = 2.25
Var[X] = sum((X - E[X])^2*P[X]) = sum(X^2*P[X]) - E[X]^2 = = 16*0.2 + 9*0.25 + 4*0 .3 + 1*0.1 + 0*0.15 - 2.25^2 = = 6.75 - 5.0625 = 1.6875 StDev[X] = sqrt(Var[X]) = 1.3. i could of messed up i did this a few times on paper and on mathway
you forgot to divide by N ?
Also, you didn't distribute the sum.. \(\sum \mu^2 = n\mu^2 \)
i'd say, use the formula you attached....and not what u have done
yes, I get 1.299 for sigma
http://www.wolframalpha.com/input/?i=%28.2*4^2%2B.25*3^2%2B.3*2^2%2B.1*1^2%2B.15*0^2%29-2.25^2
sqrt( sum ( (x - mu)^2 * P(x) ))
i need to refresh on this topic :P
that simplifies for equal probability to sqrt( sum ( (x-mu)^2 * 1/n) ) or \[ \sqrt{ \frac{1}{n} \sum (x_i-\mu)^2 }\]
ok so i am also right on C? thakn you guys for helping me
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