6) Multiple choice: Answer each sub-question by stating whether the term indicat
ID: 2929387 • Letter: 6
Question
6) Multiple choice: Answer each sub-question by stating whether the term indicated would increase, decrease, stay the same, or not enough info to say. (Note: "increase" and "decrease" refer to the absolute value.) a) In the binomial distribution, as N decreases, what happens to the value of the most likely outcome when P = .50? b) For any N, as P increases from .10 to .50, what happens to the value of the most likely outcome? c) For any N, as P increases from .50 to .90, what happens to the value of the most likely outcome? d) When P = .5, what happens to the probability of the most likely individual outcome, as N increases? 2 es? 7)Explanation / Answer
a. Since p = 0.50, 1 - p = 0.50 and the p and q terms will be (0.50)n.
The other term will be nC(n/2) for the mean.
Thus the probability of the most expected value = nC(n/2) * (0.50)n
As n decreases, nC(n/2) * (0.50)n increases.
Example: 4C2 * 0.504 = 0.375
2C1 * 0.502 = 0.5.
b) The most likely result is np.
As p increases, 1 - p which is a bigger value decreases. The net effect is that the probability of most likely value decreases.
E.g p = 0.1, n = 10 => 10 * 0.11 * 0.99 = 0.387
p = 0.2, n = 10 => 45 * 0.22 * 0.88 = 0.3020
c) The most likely result is np.
As p increases, 1 - p which is a smaller value decreases. The net effect is that the probability of most likely value increases.
E.g p = 0.8, n = 10 => 45 * 0.88 * 0.22 = 0.3020
p = 0.9, n = 10 => 10 * 0.99 * 0.11 = 0.387
d) When P = 0.5, the probability of the most likely outcome
= nC(n/2) * (0.5)n
As n increases, nC(n/2) * (0.50)n decreases.
Example: 2C1 * 0.502 = 0.5.
4C2 * 0.504 = 0.375
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