MTH 115-01 Stats Ch 6-7 Homework Name____________ Fall 2015

MTH 115-01 Stats Ch 6-7 homework Name____________ Fall 2015

For problems 1 - 4, draw the necessary pictures so that I may follow your work. Label all z-scores and areas. Assume that we have a normal distribution with the given data.

Part A: For each problem, calculate the probability as specified. (Each problem worth 5 points)

  1. P(x
  2. P(73
  3. P(x 75) = __________
  4. P(x = 73) = ___________

Part B: Determine whether each statement is true or false. If the statement is false, explain why. (Each statement worth 3 points, total 15 points)

  1. The total area under the normal distribution bell-shaped curve is infinite.
  2. The standard normal distribution is a discrete distribution.
  3. The z value corresponding to a number below the mean is always negative.
  4. The area under the standard normal distribution to the left of z = 0 is negative.
  5. For a standard normal probability distribution, the mean is always 1.

Part C: Work the following problems, showing your work for full credit. (Total 40 points)

  1. (15 points) A survey of 30 high school students about after-school work hours found a mean of 26 hours with a standard deviation of 6 hours, assuming normal distribution.

    • a. Find the best point estimate of the population standard deviation.
    • b. Find a 95% confidence interval for the population standard deviation.
  2. (10 points) To determine the number of test rides needed for a survey with 90% confidence and a margin of error of no more than 5 percentage points regarding willingness to buy a Corvette.

  3. (10 points) Of 2590 students randomly surveyed, 98% own computers. Construct a 99% confidence interval for the true proportion of all students who own computers.

  4. (15 points) A cereal packaging company fills sacks with a normally distributed mean weight of 2 pounds and a standard deviation of 0.20 pounds.

    • a) What is the probability that a randomly selected sack exceeds 2.05 pounds?
    • b) What is the probability that the mean weight of a sample of 16 sacks exceeds 2.05 pounds?

Paper For Above instruction

The following analysis addresses the key statistical concepts presented in the homework assignment, illustrating the application of normal distribution, confidence intervals, and probability calculations within the context of real-world data. The problem set reinforces understanding of the properties and behavior of distributions, as well as inferential statistics, critical skills for analyzing data in various fields.

Part A: Probability Calculations Using the Normal Distribution

Given a normal distribution, calculating the probability for specific x-values involves transforming these points into z-scores, which standardize data points relative to the mean and standard deviation. Although the problem statement does not specify the mean and standard deviation for the distribution, typical assumptions are made based on context or prior information.

For instance, suppose the distribution has a mean (μ) and standard deviation (σ). To find P(x

P(73 75), the total combined area under the curve outside that region is obtained by summing the probabilities of each tail.

Note that P(x = 73) in a continuous distribution like the normal distribution is theoretically zero, as the probability at a single point is zero.

Part B: True or False Statements on Distribution Properties

The area under the normal distribution curve extends infinitely in both directions, but the total area (probability) under the curve is exactly 1, not infinite. This statement is false because the total area is finite and equals 1.

The standard normal distribution is continuous, not discrete, characterized by a mean of 0 and a standard deviation of 1. Therefore, this statement is false.

The z-value corresponding to a point below the mean always has a negative value, reflecting its position on the left side of the mean on the standard normal curve. This statement is true.

The area to the left of z = 0 in the standard normal distribution is 0.5, representing half the total probability. Since this is a probability measure, the claim that it is negative is false.

In a standard normal distribution, the mean is always 0, not 1; the value 1 pertains to the standard deviation. Thus, this statement is false.

Part C: Applied Problems and Inferential Statistics

1. Confidence Interval for Population Standard Deviation

The problem involves 30 students with a sample mean of 26 hours and a standard deviation of 6 hours. To estimate the population standard deviation, the sample standard deviation is used directly. The best point estimate for the population standard deviation is the sample standard deviation of 6 hours.

To construct a 95% confidence interval for the population standard deviation, the chi-square distribution is used:

CI: \(\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{upper}}} , \sqrt{\frac{(n-1)s^2}{\chi^2_{lower}}} \right)\)

where \(n = 30\), \(s = 6\), and the chi-square critical values are based on degrees of freedom \(df = n-1 = 29\). Using chi-square tables, the critical values for 95% confidence are approximately 16.047 and 45.722, leading to an interval for the population standard deviation.

Calculations show the confidence interval ranges approximately from 4.04 to 9.58 hours, indicating the margin of variability within the population standard deviation.

2. Determining Sample Size for Proportion Estimate

Given the goal is to estimate the proportion of drivers willing to buy a Corvette within ±5% at 90% confidence, the sample size n can be found using the formula:

\( n = \frac{Z^2 p (1-p)}{E^2} \)

Assuming p≈0.5 for maximum variability, Z for 90% confidence is approximately 1.645, and E = 0.05. Substituting these yields n ≈ 271, indicating that approximately 271 riders need to be surveyed.

3. Confidence Interval for Population Proportion

For the survey of 2590 students with 98% owning computers, the sample proportion p̂ = 0.98. The standard error (SE) is calculated as:

SE = √[p̂(1 - p̂)/n] ≈ √[0.98×0.02/2590] ≈ 0.00088

Using a z-value of 2.576 for a 99% confidence interval, the margin of error is approximately 0.0023. The confidence interval is then approximately (0.977, 0.983), indicating high confidence that the true proportion lies within this narrow range.

4. Probability Calculations for Sack Weights

a) The probability that a sack exceeds 2.05 pounds is found using the z-score:

z = (2.05 - 2) / 0.20 = 0.25. The probability that a sack exceeds this weight is 1 - P(z

b) For a sample mean of 16 sacks, the standard error is:

SE = 0.20 / √16 = 0.05. The z-score for the sample mean exceeding 2.05 pounds is:

z = (2.05 - 2) / 0.05 = 1.0. The probability of observing a sample mean greater than 2.05 is 1 - P(z

This demonstrates how sample size affects the variability and probability of observing sample means exceeding certain thresholds.

Conclusion

This comprehensive analysis combines probability theory and inferential statistics to address real-world problems, illustrating the application of normal distribution properties, confidence interval construction, and sample size determination. Mastery of these concepts enables effective data analysis and informed decision-making across diverse disciplines.

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