For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Answer:
The probability is 0.0052
Step-by-step explanation:
Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:
P(A/B) = P(A∩B)/P(B)
The probability P(B) that at least three are aces is the sum of the following probabilities:
- The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
- There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

So, the number of ways to select exactly 3 aces is:

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725
Then, the probability P(B) that at least three are aces is:

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:
P(A∩B) = 1/270,725
Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

If <em>a</em> is fixed and <em>b</em>,<em>c</em> are unknowns then the equation <em>b</em>+<em>c</em>=10-<em>a</em> has 11-<em>a</em> solutions. They are pairs (b,c): (0,10-a), (1,9-a), (2,8-a), ... (10-a,0). As <em>a</em> runs from 0 to 10 we have total number of solutions (11-0)+(11-1)+...(11-1)=11+10+...+1=(1+11)*11/2=66.
H is less than or equal to 35
Just simplify each side and solve for the variable.
x + 4 = 15
- 4 -4
x = 11
You would subtract 4 because you need to isolate the x to get the value of x. You would subtract it to both sides because whatever you do on one side, you have to do it to the other side.