cheat at a dice game. The effect is that the 1 and 6 faces have a different probability of occurring than the 2, 3, 4 and 5 faces. Let 0 be the probability of obtaining a 1 on this biased die. Then the outcomes of rolling the biased die have the following probability mass function. Table 1 The p.m.f. of outcomes of rolls of a biased die Outcome 2. 3. 4. Probability 0 1–20) (1- 20) {1-20) (1- 20) 0 |(1 – 1= 20) 0 By consideration of the p.m.f. in Table 1, explain why it is necessary for 0 to be such that 0 < 0 < 1/2. (ii) The value of 0 is unknown. Data from which to estimate the value of 0 were obtained by rolling the biased die 1000 tnes. The result of this experiment is shown in Table 2. (a) Let X and Y be independent random variables both with the same mean ji 7 0. Define a new random variable W= aX + bY, where a and b are constants.

MATLAB: An Introduction with Applications
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cheat at a dice game. The effect is that the 1 and 6 faces have a
different probability of occurring than the 2, 3, 4 and 5 faces.
Let 0 be the probability of obtaining a 1 on this biased die. Then the
outcomes of rolling the biased die have the following probability mass
function.
Table 1
The p.m.f. of outcomes of rolls of a biased die
Outcome
2.
3.
6.
Probability 0 4(1–20) (1- 20) {1- 20) (1= 20) 0
(i) By eonsidđeration of the p.m.f. in Table 1, explain why it is
necessary for 0 to be such that 0 < 0 < /2.
(ii) The value of 0 is unknown. Data Trom which to estimate the value
of 0 were obtained by rolling the biased die 1000 times. The résult
of this experiment is shown in Table 2.
(a) Let X and Y be independent random variables both with the same
mean ji 7 0. Define a new random variable W aX + 6Y, where a and
b are constants.
(1) Obtain an expression for E(W).
(i) What eonstraint is there on the values of a and b so that W is an
unbiased estimator of µ? Hence write all unbiased versions of W as
a formula involving a, X and Y only (and not b).
Accessibility: Investigate
DFocur
BA
日
Transcribed Image Text:cheat at a dice game. The effect is that the 1 and 6 faces have a different probability of occurring than the 2, 3, 4 and 5 faces. Let 0 be the probability of obtaining a 1 on this biased die. Then the outcomes of rolling the biased die have the following probability mass function. Table 1 The p.m.f. of outcomes of rolls of a biased die Outcome 2. 3. 6. Probability 0 4(1–20) (1- 20) {1- 20) (1= 20) 0 (i) By eonsidđeration of the p.m.f. in Table 1, explain why it is necessary for 0 to be such that 0 < 0 < /2. (ii) The value of 0 is unknown. Data Trom which to estimate the value of 0 were obtained by rolling the biased die 1000 times. The résult of this experiment is shown in Table 2. (a) Let X and Y be independent random variables both with the same mean ji 7 0. Define a new random variable W aX + 6Y, where a and b are constants. (1) Obtain an expression for E(W). (i) What eonstraint is there on the values of a and b so that W is an unbiased estimator of µ? Hence write all unbiased versions of W as a formula involving a, X and Y only (and not b). Accessibility: Investigate DFocur BA 日
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