Please help me figure them out.

1. In the 105th Congress, the Senate consist of 55 Republicans and 45 Democrats. If a committee is formed by randomly selecting four different senators, what is the probability that they're all Republicans?

2. It's common for public opion polls to have a confidence level of 95% meaning that there is .95 probability that the poll results are accurate within margin of error. If five different organizations conduct independent polls, whats the probability that all five of them are accurate within the claimed margins of error?

3. Determine if property distributions is givin. If not, identify requirements not satisfied. If it is, find mean and standard deviation.

In a study of brand recognition of Sony, groups of four consumers are interviewd. If x is the number in the group who recognize the Sony brand name, then x can be 0, 1, 2, 3, or 4 and the corresponding probabilities are 0.0016, 0.0250, 0.1432, 0.3892, and 0.4096.

4. Ten percent of American adults are left handed. A statistics class has 25 students in attendance.

a. find the mean and standard deviation for the number of left handed students in such classes of 25 studnets.
b. would it be unusual to survey a class of 25 stduents and find 5 of them are lefthanded. why?

1) (55/100)*(54/99)*(53/98)*(52/97)

2) .95^5

3) The sum of the probabilities should add to 1

4) mean of binomial is n*p, standard deviation is sqrt(n*p*(1-p)). Take it from here.

1. To find the probability that all four randomly selected senators are Republicans, you can use the principle of independent events. The probability of selecting a Republican senator on the first draw is 55/100 (since there are 55 Republicans out of 100 senators). After one Republican senator is selected, there are now 54 Republicans left out of 99 senators. So the probability of selecting another Republican on the second draw is 54/99. Similarly, for the third and fourth draws, the probabilities would be 53/98 and 52/97, respectively. To find the probability of all four events happening, you multiply these probabilities together:

P(all Republicans) = (55/100) * (54/99) * (53/98) * (52/97)

2. The probability that each individual poll is accurate within the claimed margin of error is 0.95. Since each poll is conducted independently, you can multiply the probabilities together to find the probability that all five polls are accurate within the claimed margins of error:

P(all accurate) = (0.95)^5

3. To determine if the probability distribution is given, you need to check if the sum of the probabilities for each possible value of x (0, 1, 2, 3, and 4) adds up to 1. In this case, the sum of the probabilities provided is:

0.0016 + 0.0250 + 0.1432 + 0.3892 + 0.4096 = 0.9686

Since the sum of the probabilities is not equal to 1, the probability distribution is not given. To find the mean and standard deviation, you would need the correct probabilities.

4. For a binomial distribution with a probability of success (being left-handed) of 0.10, and a sample size of 25, you can find the mean and standard deviation.

a. The mean of a binomial distribution is calculated by multiplying the sample size (n) by the probability of success (p). In this case:

Mean = n * p = 25 * 0.10 = 2.5

The standard deviation of a binomial distribution is calculated using the formula sqrt(n * p * (1 - p)). In this case:

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(25 * 0.10 * (1 - 0.10)) = sqrt(2.25) = 1.5

b. To determine if it would be unusual to find 5 left-handed students in a class of 25, you can calculate the z-score. The formula for the z-score is (x - mean) / standard deviation, where x is the observed value.

In this case, the observed value is 5, the mean is 2.5, and the standard deviation is 1.5. Calculating the z-score:

z = (5 - 2.5) / 1.5 = 1.67

The z-score tells you the number of standard deviations away from the mean the observed value is. You can then compare the z-score to determine if it is unusual. Typically, a z-score greater than 2 or less than -2 is considered unusual. In this case, a z-score of 1.67 is not considered unusual.