help please!!!

1. The average number of bumped passengers per day on FliRite Airlines is 2.47 with a standard deviation of 1.75. For a random sample of 40 days, what is the chance the average number of bumped passengers in the sample is between 2 and 2.75?
a. About 91%
b. About 23 %
c. About 80%
d. Virtually 100%

2. In a certain school district, two hundred fifth grade children are chosen at random and their heights are noted. The AVERAGE height of these children is 56 inches with a standard deviation of 4 inches. Find a 95% confidence interval for the AVERAGE height of all fifth grade children in the district.
a. 55.7" to 56.3"
b. 48" to 54"
c. 52" to 60"
d. 55.4" to 56.6"

3. A bank manager wants to know the average daily balance of all the checking accounts at her branch. She randomly samples 144 checking accounts. The average daily balance of the checking accounts in the sample is $280. The SD is $66. Find an approximate 95% confidence interval for the average daily balance of all checking accounts at this branch.
a. $269 to $291
b. $148 to $412
c. $279 to $281
d. $275 to $285

4. A survey organization takes a simple random sample of 725 households from a city of 90,000 households. On the average, there are 2.45 persons per sample household, and the SD is 1.65. Choose the statement below that is FALSE.
a. The SE for the sample average is 0.06.
b. A 95% confidence interval for the average household size in the city is approximately 2.33 to 2.57.
c. 95% of the households in the city contain between 2.33 and 2.57 persons.
d. The 95% confidence interval is about right because, with 725 draws from the box, the probability histogram for the average of the draws follows the normal curve.

5. The salaries of a random sample of 288 Nevada teachers produced a 90% confidence interval for the average salary of $38,944 to $42,893. Choose which statement below is TRUE.
a. If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce this confidence interval.
b. If we took many random samples of Nevada teachers, about 90% of them would produce a confidence interval that contained the mean salary of all Nevada teachers.
c. About 90% of Nevada teachers earn between $38,944 and $42,893.
d. About 90% of the teachers surveyed earned between $38,944 and $42,893.
e. We are 90% confident that the average teacher salary in the United States is between $38,944 and $42,893.

1. C

2. D
3. A

1. To find the probability that the average number of bumped passengers in the sample is between 2 and 2.75, we need to use the Central Limit Theorem, which states that the distribution of sample means will approximate a normal distribution regardless of the shape of the population distribution, as long as the sample size is large enough.

Since the sample size is 40, which is reasonably large, we can assume that the distribution of sample means will be approximately normal.

First, we need to calculate the standard error of the sample mean. The formula for standard error (SE) is:

SE = standard deviation / square root of sample size

SE = 1.75 / sqrt(40) = 0.2765

To transform the problem into a standard normal distribution, we calculate the z-scores for both 2 and 2.75, using the formula:

z = (x - mean) / SE

z1 = (2 - 2.47) / 0.2765 ≈ -1.66
z2 = (2.75 - 2.47) / 0.2765 ≈ 1.01

Next, we look up the probabilities corresponding to these z-scores in a standard normal distribution table or use statistical software.

P(-1.66 < Z < 1.01) ≈ 0.9105 - 0.4515 ≈ 0.459

So, the chance that the average number of bumped passengers in the sample is between 2 and 2.75 is approximately 45.9%. Therefore, the correct answer is NOT provided in the options.

2. To find a 95% confidence interval for the average height of all fifth grade children in the district, we can use the formula:

CI = (mean - margin of error, mean + margin of error)

The margin of error can be calculated using the formula:

ME = critical value * standard deviation / square root of sample size

For a 95% confidence interval, the critical value corresponds to a z-score of ±1.96 (approximately).

ME = 1.96 * 4 / sqrt(200) ≈ 1.746

Therefore, the 95% confidence interval for the average height of all fifth-grade children in the district is:

56 - 1.746 to 56 + 1.746 = 54.254 to 57.746

So, the correct answer is NOT provided in the options.

3. To find an approximate 95% confidence interval for the average daily balance of all checking accounts at the branch, we can use the same formula as in question 2:

CI = (mean - ME, mean + ME)

The margin of error can be calculated using the formula:

ME = critical value * standard deviation / square root of sample size

For a 95% confidence interval, the critical value corresponds to a z-score of ±1.96 (approximately).

ME = 1.96 * 66 / sqrt(144) ≈ 15.50

Therefore, the 95% confidence interval for the average daily balance of all checking accounts at this branch is:

280 - 15.50 to 280 + 15.50 = 264.50 to 295.50

So, the correct answer is provided in option a. Approximately $269 to $291.

4. The statement that is FALSE is:

d. The 95% confidence interval is about right because, with 725 draws from the box, the probability histogram for the average of the draws follows the normal curve.

The reason this statement is false is that the Central Limit Theorem states that the distribution of sample means will approximate a normal distribution when the sample size is large enough (usually considered to be at least 30). In this case, the sample size is 725, which is large enough to assume a normal distribution for the sample mean.

5. The statement that is TRUE is:

b. If we took many random samples of Nevada teachers, about 90% of them would produce a confidence interval that contained the mean salary of all Nevada teachers.

A 90% confidence interval represents the range within which the true population parameter (average salary of all Nevada teachers) is expected to fall 90% of the time if we repeatedly sample and compute confidence intervals. Therefore, about 90% of the confidence intervals constructed from these repeated samples would contain the true mean salary.