Question

Let Y be a binomial random variable with n = 10 and p = .2. a Use Table 1, Appendix 3, to obtain P(2 < Y < 5) and P(2 ≤ Y < 5). Are the probabilities that Y falls in the intevals (2, 5) and [2, 5) equal? Why or why not? b Use Table 1, Appendix 3, to obtain P(2 < Y ≤ 5) and P(2 ≤ Y ≤ 5). Are these two probabilities equal? Why or why not? c Earlier in this section, we argued that if Y is continuous and a < b, then P(a < Y < b) = P(a ≤ Y < b). Does the result in part (a) contradict this claim? Why?

Answer 1

In a binomial distribution with n = 10 and p = 0.2, P(2 < Y < 5) and P(2 ≤ Y < 5) are not equal because Y = 2 is included in the latter. P(2 < Y ≤ 5) and P(2 ≤ Y ≤ 5) are equal because adding Y = 2, which has a probability of zero, does not change the outcome. This does not contradict the properties of continuous random variables as the binomial is discrete.

Step-by-step explanation:

For a binomial random variable Y with n = 10 and p = 0.2, the probability of Y taking on a value between 2 and 5 can be found using a binomial table or computational tools such as a calculator. The probability P(2 < Y < 5) is obtained by summing the probabilities for Y = 3 and Y = 4, as Y cannot take on non-integer values. Similarly, for P(2 ≤ Y < 5), we also include Y = 2 in our sum.

To find P(2 < Y ≤ 5), we would use Y = 3, Y = 4, and Y = 5. For P(2 ≤ Y ≤ 5), Y = 2 is additionally included. The probabilities P(2 < Y < 5) and P(2 ≤ Y < 5) are not equal, because the second includes the probability of Y = 2. However, P(2 < Y ≤ 5) and P(2 ≤ Y ≤ 5) are equal in the binomial case, because adding the zero-probability event Y = 2 to P(2 < Y ≤ 5) does not change its value.

In the context of continuous random variables, the probability of any single value (e.g., P(Y = c)) is 0. Therefore, for a continuous variable, the probabilities P(a < Y < b) and P(a ≤ Y < b) are equivalent, as including the endpoints does not affect the total probability. This concept does not contradict part (a), as the binomial distribution is discrete, and each integer value of Y has a non-zero probability.

Answer 2

a. (P(2 < Y < 5) = 0.8) and (P(2 ≤ Y < 5) = 0.9). The probabilities that Y falls in the intervals (2, 5) and [2, 5) are not equal.

b. (P(2 < Y ≤ 5) = 0.9) and (P(2 ≤ Y ≤ 5) = 0.7). These two probabilities are not equal.

c. The result in part (a) does not contradict the claim. The distinction between (<) and
`(\leq\)` in probability intervals is significant, and the probabilities calculated based on the table are consistent with this distinction.

Step-by-step explanation:

a. Using Table 1, Appendix 3, we find
`\(P(2 < Y < 5) = P(Y = 3) + P(Y = 4) = 0.3 + 0.5 = 0.8\) and \(P(2 ≤ Y < 5) = P(Y = 2) + P(Y = 3) + P(Y = 4) = 0.1 + 0.3 + 0.5 = 0.9\).`

b. Similarly, (P(2 < Y ≤ 5) = P(Y = 3) + P(Y = 4) + P(Y = 5) = 0.3 + 0.5 + 0.1 = 0.9), and (P(2 ≤ Y ≤ 5) = P(Y = 2) + P(Y = 3) + P(Y = 4) + P(Y = 5) = 0.1 + 0.3 + 0.5 + 0.1 = 0.7).

c. The result in part (a) does not contradict the claim because the probabilities are calculated consistently with the definitions of the intervals. The difference lies in the inclusion or exclusion of the endpoint, and the calculated probabilities align with this distinction in probability intervals.