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vovikov84 [41]
3 years ago
10

A die is rolled. If you roll a 1, 2 or 3, you will toss 10 coins. If you roll a 4, 5 or 6, you will toss 20 coins. Let X denote

the number of heads tossed. a) Determine the support of X. b) Determine the mean of X. c) Determine the variance of X.
Mathematics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

a) The support of X is {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

b) The mean of X is 7.5

c) The variance of X is 10

Step-by-step explanation:

a) Since you can toss up to 20 coins, and from that you can obtain any number of heads from 0 to 20, then the support of X {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}.

b) To compute the mean of X, we need to see how is the <em>behavior</em> of X. Since 3 dices will gives 10 coins to toss and the other 3 will give us 20, then there is a probability of 1/2 that we tossed 10 coins and a probability of 1/2 that we tossed 20.

The random variable X, conditioned to the event 'The dice is 1,2 or 3' (or equivalently, 10 coins are tossed), will have a binomial distribution with paramenters n = 10, p = 1/2. The mean of X in this case is np = 5. If X is conditioned to the event 'The dice is 4,5 or 6', then X will have also binomial distribution, but this time with paramenters n = 20, p = 1/2. The mean of X in this case is 20*1/2 = 10.

Since each event we conditioned in had probability 1/2 to occur, then E(X) = 1/2 * 5 + 1/2 * 10 = 7.5.

c) Remember that V(X) = E(X²)- E(X)². Since we alredy know the mean of X, we just need to compute the mean of X squared.

The variance of a binomial distribution Z with paramenters n and p is

V(Z) = np(1-p)

since V(Z) = E(Z²)- E(Z)² = E(Z²)- n²p², then

E(Z²) = V(Z) + E(Z)² = np(1-p) + n²p² = p²(n²-n) + np

Therefore

E(X²) = E(X² | 10 coins are tossed) * 1/2 + E(X² | 20 coins are tossed) * 1/2 =

1/2*(0.5²(10²-10) + 5) + 1/2*(0.5²(20²-20) + 10) = 13.75 + 52.5 = 66.25

As a consecuence

V(X) = 66.25 - 7.5² = 10

The variance of X is 10.

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Answer:

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Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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