Answer:
2.5th percentile and the 97.5th percentile.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
So we obtain the 0.025*100 = 2.5th percentile and the (1-0.025)*100 = 97.5th percentile.
So the answer is:
2.5th percentile and the 97.5th percentile.
First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of
possible passwords.
Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of
possible passwords.
So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is
Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have
As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:
Where x can be any of the 36 characters.
So, we have 25 cases with 4 vacant slots, leading to a probability of
Finally, you can compute the probability of the union using the formula
Since we already computed all these quantities.