Statistics-mean-standard deviation?

A huge vat is full of marbles all mixed together, with a number written on it. 20% of the marbles have a "0", 30% have a "1", 10% have a "15", 10% have a "30", 10% have a "80" , and the rest have a"100". Let Y be a number written on a randomly-selected marble (round all answers to four decimal places)





[a]The mean of the random variable Y is:?





[b]The Standard Dev. of Y is:?





[c]Take 20 marbles. What is the probability that the number of "ones" is at most 5:?





[d]Take 100 Marbles. What is the prob. that the number of "ones" is at most 25:?





[e]The prob of the average of numbers on n=25 is less than 35 is:?


For a sample of n=81 it would be:?


For a sample of n=400 it would be:?


For a sample of n=1600 it would be:?


For a sample of n=6400 it would be:?

Statistics-mean-standard deviation?
a) mean is just a weighted average:


m = .20(0) + .30(1) + .1(15) + .1(30) + .1(80) +.2(100) = 32.8





b) standard deviation:


SD = sqrt[.20((-32.8)^2) + .30((-31.8)^2) + .10((-17.8)^2) + .10((-2.8)^2) + .10(47.2^2) +.20(67.2^2)] = 39.0629





c) Since it's a "huge vat", we'll assume that taking out 20 marbles doesn't affect the proportions of the different kinds of marbles in the vat. The probability of exactly





zero "ones" = (1-.3)^20 = 0.000797923


one "one" = 20(.3*.7^19) = 0.006839337


two "ones" = (20*19)(.3^2*.7^18)/2 = 0.027845873


three "ones" = (20*19*18)(.3^3*.7^17)/6 = 0.071603672


four "ones" = ... = 0.130420974


five "ones" = ... = 0.178863051





Adding these up gives 0.416370829 as the probability of getting at most five "ones".





d) Exact same logic as before gives probability of 0.163130104 for getting at most twenty-five "ones".





e) sorry, no time to work on this last one -- you'll have to find someone else.
Reply:....0 | 0.20 |.. 0.0 | 215.168


....1 | 0.30 |.. 0.3 | 303.372


..15 | 0.10 |.. 1.5 |.. 31.684


..30 | 0.10 |.. 3.0 |.... 0.784


..80 | 0.10 |.. 8.0 | 222.784


100 | 0.20 | 20.0 | 903.168


[a] μ = 32.8


[b] σ ≈ 40.95070


[c] 1 - 0.7^15 ≈ 0.9952524


[d] 1 - 0.7^75 ≈ 1.0


[e] Z = 5(35 - 32.8) / 40.95070 = 0.2686157


P(m %26lt; 35) = 0.605889


etc.
Reply:let Y=number written


y ---- p(y)


0-----0.20


1-----0.30


15----0.10


30----0.10


80----0.10


100--0.20


mean= E(y) = sum y(i).p(y(i) for each i


0(0.20)+1(0.30)+15(0.10)+30(0.10)+80(0...


b) find variance of y


[y-E(y)]^2


[0-32.8]^2=1075.84


[1-32.8]^2=1011.24


..


..


[100-32.8]^2=4515.84


add all (y-E(y))^2 and take its square root.


It's your standard deviation.


c)prob (y=1)=0.30


Use a Binomial distribution with n=20 p=0.3


x=0,1,2,3,4,5 and sum these values


Use the source if you don't have a Binomial table.=0.4164


d)n=100 p=0.3 x=0 through 25.


=0.1631


e) The question is not clear. What is your definition of n? I have defined n as my number of marbles. The probability of a marble showing less than 35 is 0.70 from the table.


average of numbers is 32.8 .


p(av %26lt; 35)=p(z %26lt; (35-32.8)/sd/sqrt(25)) -- (1)assuming a normal distribution)


sd can be computed from the table.


for a sample of n=81


replace 25 by 81 in (1)


and so on. Use the normal table to evaluate these probabilities.