Miami Dade College 2023 Statistical Methods Q&A

Miami Dade Collegesta 2023 Statistical Methodsuse The Following Dis

Use the following distribution to answer questions 3 and 4. What is the total area under the curve? A) 1 B) 3 C) 2500 D) What is the mean? A) 1 B) 3 C) 2500 D) 3000. Find the area under the standard normal distribution curve. Between z = 0 and z = .75 A) .07734 B) .7734 C) .2734 D) To the left of z = -1.39 A) .0823 B) .0082 C) .82 D) To the left of z = 1.23 A) .8907 B) .1093 C) .2266 D) To the right of z = 1.23 A) .1093 B) .8907 C) .2266 D) Between z = -0.75 and z = 1.23 A) .8907 B) .1093 C) .2266 D) .6641

The average credit card debt for college seniors is $3,262 with a standard deviation of $1,100. Find these probabilities: (1) That a senior owes at least $1,000, (2) that a senior owes more than $4,000. (3) The debt is normally distributed, so z-scores need to be calculated for these amounts.

The test scores for police academy qualifiers are normally distributed, with a mean of 200 and a standard deviation of 20. To qualify, candidates must score in the top 10%. Find the minimum score required on this test for qualification.

The average credit card debt for college seniors, now at $3,173 with a standard deviation of $1,120, is analyzed to find the probability that a randomly selected senior has a balance less than $2,700, and the probability that the mean balance of a sample of students is less than $2,700.

In a binomial experiment with N=45, p=0.47, and q=0.53, find the mean and standard deviation of the number of successes.

A study of 35 ninth-grade students reports a mean of 16.6 hours per week of video game play with a standard deviation of 2.8 hours. Construct the 95% confidence interval for the population mean time.

A scientist estimates the average river depth with 99% confidence that the error margin is within 2 feet, given previous data with a standard deviation of 4.33 feet. Calculate the minimum sample size needed.

Ten people are surveyed on their sleep duration with a mean of 7.1 hours and a standard deviation of 0.78 hours. Find the 95% confidence interval for the population mean sleep time.

A survey of 1500 adults indicates that 39% plan to take more vacations. Find the 95% confidence interval for the proportion of all adults who will take more vacations.

An investigator wants to estimate the proportion of home computer owners with 95% confidence and a margin of error of 2%, based on an initial estimate that 40% own a computer. Find the minimum sample size needed.

Paper For Above instruction

The distribution's total area under the curve is a fundamental concept in understanding probability distributions, especially the normal distribution. The correct answer is A) 1, which signifies that the total probability over the entire distribution sums to one, reflecting the certainty that a randomly selected value falls somewhere under the curve.

The mean of a distribution is a measure of its central tendency, representing the average of all possible values weighted by their probabilities. Based on the options, the most appropriate answer is B) 3, aligning with the context provided in the dataset or distribution parameters. In a standard normal distribution, the mean is expected to be zero, but given specific data points, an alternative mean might be provided.

Finding the area under the standard normal curve between two z-scores involves using z-tables or statistical software. Between z=0 and z=0.75, the cumulative probability is approximately 0.2734 (option C), indicating the area between these z-scores under the standard normal curve.

To the left of z=-1.39, the cumulative probability is about 0.0823 (option B). This reflects the proportion of data below this z-score, essential for percentile or percentile rank computations. Similarly, for z=1.23, the cumulative probability is approximately 0.8907 (option A), indicating the proportion of data below this z-score.

For z=1.23, the area to the right involves subtracting the cumulative probability from one: 1 - 0.8907 = 0.1093 (option B). For the combined range between z=-0.75 and z=1.23, the area is approximately 0.6641 (option D), which can be obtained by subtracting the cumulative probabilities at these z-scores.

The average credit card debt among college seniors is approximately $3,262 with a standard deviation of $1,100. To find the probability that a senior owes at least $1,000, we calculate a z-score: (1000-3262)/1100 ≈ -2.52, corresponding to a cumulative probability near 0.0059, hence a probability of about 0.9941 that debt is at least $1,000. Similarly, for debt exceeding $4,000, z = (4000-3262)/1100 ≈ 0.679, with an associated probability of about 0.249, so approximately 75.1% owe more than $4,000.

For police test scores, the upper percentile score corresponds to the top 10%, thus the 90th percentile. The z-score associated with the 90th percentile is approximately 1.28. Using the formula x = μ + zσ, the minimum qualifying score is 200 + (1.28)(20) = 225.6, which rounds to 226, confirming option A.

With a mean debt of $3,173 and a standard deviation of $1,120, the probability that a randomly selected undergraduate's debt is less than $2,700 is found by computing the z-score: (2700-3173)/1120 ≈ -0.42, which has a cumulative probability of about 0.3372 (option C). For the sample mean, the standard error is σ/√n. Calculating the probability that the mean debt of a sample is less than $2,700 involves z = (2700-3173)/(1120/√n). Assuming the sample size n is large enough, the probability remains approximately 0.3372 (option C).

In binomial probability distributions, the mean number of successes is given by Np, and the standard deviation by √Np q. With N=45, p=0.47, q=0.53, the mean is 450.47=21.15 (option A), and the standard deviation is √(450.47*0.53) ≈ 3.35 (option C).

The confidence interval for the average hours spent playing video games among ninth-graders is constructed using sample mean and standard deviation, along with the t-distribution for small samples. The 95% confidence interval is calculated as mean ± t* (s/√n). With a mean of 16.6, standard deviation 2.8, and n=35, the interval approximately is (15.7, 17.5), matching option B.

The minimum sample size required for estimating the river depth within an error margin of 2 feet with 99% confidence, given a standard deviation of 4.33 feet, is calculated using the formula n = (Zσ/E)^2. With Z for 99% confidence about 2.576, n = (2.5764.33/2)^2 ≈ 31.4, rounded to 32 (option B).

For a sleep survey of 10 individuals with a mean of 7.1 hours and standard deviation of 0.78 hours, the 95% confidence interval for the population mean is computed as mean ± t* (s/√n). With degrees of freedom 9, t-value approximately 2.262, the interval is roughly (6.54, 7.66), aligning with option D.

The proportion of adults planning to take more vacations, 39%, is used to create a confidence interval for the true proportion with size n=1500, and margin of error calculated as Z*(√p(1-p)/n). With p=0.39 and Z=1.96 for 95% confidence, the interval is approximately (0.36, 0.42).

To estimate the proportion of home computer owners within 2% precision and a confidence level of 95%, the sample size is calculated using n = (Z^2 * p(1-p))/E^2. Substituting Z=1.96, p=0.40, E=0.02 yields n ≈ 2305 (option B).

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