All I need is D answered. 3. A nefarious gambler has developed a weighted six-si

Question

All I need is D answered.

3. A nefarious gambler has developed a weighted six-sided die, with sides marked 1, 2, 3, 4, 5, and 6. The probability that a 6 is rolled is 0.4, and all the other sides are equally likely. Define the random variable X as the value on the die when it is rolled once. (a) Write the pmf of X. That is, make a table of the values X can take and the associated probabilities. (b) Find the probability that an even number is rolled (c) Find the mean (or expected value) and variance of X. (d) The gambler has a second weighted die designed like the first. A game is based on the number W = 2X +Y, where X is the number that shows up on the first weighed die and Y is the number that shows up on the second i. Find the expected value (or mean) of W ii. Find the standard deviation of W

Explanation / Answer

here probability of 1,2,3,4,5 =1-probability of 6 =1-0.4 =0.6

hence probability of any of 1,2,3,4,5 =0.6/5 =0.12

from above:

therefore mean of X =E(X) =4.2

and Variance =Var(X) =3.36

d)

i) expected value of W =E(W) =E(2X+Y) =2E(X)+E(Y) =2*4.2+4.2=12.6

(ii) Std deviation of W =SD(W) =(22Var(X)+Var(Y))1/2 =(4*3.36+3.36)1/2 =4.0988

x p(x) xP(x) x2P(x) (x-)2 (x-)2P(x) 1 0.1200 0.120 0.120 10.240 1.229 2 0.1200 0.240 0.480 4.840 0.581 3 0.1200 0.360 1.080 1.440 0.173 4 0.1200 0.480 1.920 0.040 0.005 5 0.1200 0.600 3.000 0.640 0.077 6 0.4000 2.400 14.400 3.240 1.296 total 1 = 4.20 21.000 20.440 2= 3.3600 std deviation=     =    2 = 1.8330

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