1) How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) for sleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution.
50 65 100 90 90 50 30 23 100 110
105 95 105 60 110 120 95 90 60 70
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (Round your answers to two decimal places.)
(b) Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price μ of all summer sleeping bags. (Round your answers to two decimal places.)
- lower limit $
- upper limit $
2) Suppose an x distribution has mean μ = 3. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81.
(a) What is the value of the mean of each of the two x distributions?
- For n = 49, μ
- X =
- For n = 81, μ
- X =
(b) For which x distribution is P( x > 3.75) smaller? Explain your answer.
- The distribution with n = 81 because the standard deviation will be larger.
- The distribution with n = 49 because the standard deviation will be smaller.
- The distribution with n = 49 because the standard deviation will be larger.
- The distribution with n = 81 because the standard deviation will be smaller.
(c) For which x distribution is P(2.25 < x < 3.75) greater? Explain your answer.
- The distribution with n = 81 because the standard deviation will be larger.
- The distribution with n = 81 because the standard deviation will be smaller.
- The distribution with n = 49 because the standard deviation will be smaller.
- The distribution with n = 49 because the standard deviation will be larger.
3) Suppose the heights of 18-year-old men are approximately normally distributed, with mean 66 inches and standard deviation 3 inches.
(a) What is the probability that an 18-year-old man selected at random is between 65 and 67 inches tall? (Round your answer to four decimal places.)
(b) If a random sample of twenty-four 18-year-old men is selected, what is the probability that the mean height x is between 65 and 67 inches? (Round your answer to four decimal places.)
(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?
- The probability in part (b) is much higher because the mean is smaller for the x distribution.
- The probability in part (b) is much higher because the standard deviation is larger for the x distribution.
- The probability in part (b) is much higher because the mean is larger for the x distribution.
- The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.
- The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
4) Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.26 gram.
(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)
- lower limit
- upper limit
- margin of error
(b) What conditions are necessary for your calculations? (Select all that apply.)
uniform distribution of weights
- σ is known
- σ is unknown
- n is large
- normal distribution of weights
(c) Give a brief interpretation of your results in the context of this problem.
- There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
- The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.
- There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
- The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
- The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.14 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
Hummingbirds
5) At wind speeds above 1000 centimeters per second (cm/sec), significant sand-moving events begin to occur. Wind speeds below 1000 cm/sec deposit sand and wind speeds above 1000 cm/sec move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes. At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty wind speed readings gave an average velocity of x = 1075 cm/sec. Based on long-term experience, σ can be assumed to be 250 cm/sec.
(a) Find a 95% confidence interval for the population mean wind speed at this site. (Round your answers to the nearest whole number.)
- lower limit cm/sec
- upper limit cm/sec
(b) Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.
- No. This interval indicates that the population mean wind speed is such that the sand may not always be moving at this site.
- Yes. This interval indicates that the population mean wind speed is such that the sand may not always be moving at this site.
- Yes. This interval indicates that the population mean wind speed is such that the sand is always moving at this site.
- No. This interval indicates that the population mean wind speed is such that the sand is always moving at this site.
6) Suppose x has a distribution with μ = 14 and σ = 11.
(a) If a random sample of size n = 45 is drawn, find μx, σ x and P(14 ≤ x ≤ 16). (Round σx to two decimal places and the probability to four decimal places.)
- μx =
- σ x =
- P(14 ≤ x ≤ 16) =
(b) If a random sample of size n = 68 is drawn, find μx, σ x and P(14 ≤ x ≤ 16). (Round σ x to two decimal places and the probability to four decimal places.)
- μx =
- σ x =
- P(14 ≤ x ≤ 16) =
(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is part (a) because of the sample size. Therefore, the distribution about μx is .
7) Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below. Assume that the population of x values has an approximately normal distribution.
92 173 133 97 75 94 116 100 85
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean startup cost x and sample standard deviation s. (Round your answers to one decimal place.)
- x = thousand dollars
- s = thousand dollars
(b) Find a 90% confidence interval for the population average startup costs μ for candy store franchises. (Round your answers to one decimal place.)
- lower limit thousand dollars
- upper limit thousand dollars
8) Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 5. (Round your answer to three decimal places.)
9) Suppose x has a distribution with μ = 33 and σ = 8.
(a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means?
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 8.
- No, the sample size is too small.
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 2.
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 0.5.
(b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16?
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 2.
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 8.
- Yes, the x distribution is normal with mean μ x = 33 and σ x = 0.5.
- No, the sample size is too small.
Find P(29 ≤ x ≤ 34). (Round your answer to four decimal places.)
10) Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean μ = 6600 and estimated standard deviation σ = 2500. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.
(a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)
(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x?
- The probability distribution of x is approximately normal with μx = 6600 and σx = 1250.00.
- The probability distribution of x is approximately normal with μx = 6600 and σx = 1767.77.
- The probability distribution of x is not normal.
- The probability distribution of x is approximately normal with μx = 6600 and σx = 2500.
What is the probability of x < 3500? (Round your answer to four decimal places.)
(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)
(d) Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased?
- The probabilities increased as n increased.
- The probabilities stayed the same as n increased.
- The probabilities decreased as n increased.
- If a person had x < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse?
- It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.
- It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia.
- It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.
- It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia.