Introduction to Statistical Thinking

Consult Table 4.6.1. The probabilities of the different values of \(Y\) are \(\{p, 2p, \ldots, 6p\}\text{.}\) These probabilities sum to 1, consequently

The event \(\{Y < 3\}\) contains the values 0, 1 and 2. Therefore,

The event \(\{Y = \mbox{odd}\}\) contains the values 1, 3 and 5. Therefore,

The event \(\{1 \leq Y < 4\}\) contains the values 1, 2 and 3. Therefore,

The event \(\{|Y -3| < 1.5\}\) contains the values 2, 3 and 4. Therefore,

The values that the random variable \(Y\) obtains are the numbers 0, 1, 2, …, 5, with probabilities \(\{1/21, 2/21, \ldots, 6/21\}\text{,}\) respectively. The expectation is obtained by the multiplication of the values by their respective probabilities and the summation of the products. Let us carry out the computation in R:

Y.val <- c(0,1,2,3,4,5)
P.val <- c(1,2,3,4,5,6)/21
E <- sum(Y.val*P.val)
E

We obtain an expectation \(\Expec(Y) = 3.3333\text{.}\)

The values that the random variable \(Y\) obtains are the numbers 0, 1, 2, …, 5, with probabilities \(\{1/21, 2/21, \ldots, 6/21\}\text{,}\) respectively. The expectation is equal to \(\Expec(Y) = 3.333333\text{.}\) The variance is obtained by the multiplication of the squared deviation from the expectation of the values by their respective probabilities and the summation of the products. Let us carry out the computation in R:

Var <- sum((Y.val-E)^2*P.val)

We obtain a variance \(\Var(Y) = 2.2222\text{.}\)

The standard deviation is the square root of the variance: \(\sqrt{\Var(Y)} = \sqrt{2.2222} = 1.4907\text{.}\)

An outcome of the game of chance may be represented by a sequence of length three composed of the letters “H" and “T". For example, the sequence “THH“ corresponds to the case where the first toss produced a ”Tail", the second a “Head" and the third a “Head".

With this notation we obtain that the possible outcomes of the game are \(\{\mbox{HHH}, \mbox{THH},\mbox{HTH}, \mbox{TTH},\mbox{HHT}, \mbox{THT},\mbox{HTT}, \mbox{TTT}\}\text{.}\) All outcomes are equally likely. There are 8 possible outcomes and only one of which corresponds to winning. Consequently, the probability of winning is 1/8.

Consider the previous solution. One looses if any other of the outcomes occurs. Hence, the probability of loosing is 7/8.

Denote the gain of the player by \(X\text{.}\) The random variable \(X\) may obtain two values: 10-2 = 8 if the player wins and -2 if the player looses. The probabilities of these values are {1/8, 7/8}, respectively. Therefore, the expected gain, the expectation of \(X\) is: