Download Lars Hertel STATS 7 Discussion 5 Let the random variable X

Lars Hertel
STATS 7
Discussion 5
1
Let the random variable X represent the weights, in pounds (lb), of adult male grizzly bears. X can be
modeled using a normal distribution with a mean of 845 lb and a standard deviation of 35 lb. For questions requiring a calculator, be sure to show what you typed in your calculator along with your answers.
a. Calculate P (800 ≤ X ≤ 900). See slide 7 of Continuous random variable notes.
b. What is the probability that an adult male grizzly bear weighs either less than 800 lb or more than
900 lb?
c. What is the value of the first quartile of weights in the population of adult male grizzly bears? See
slide 9.
d. What is the range of mean weights in random samples of 4 adult male grizzly bears that represents
the middle 99.7% of all possible mean weights? Sketch your answer on a normal curve.
e. Go to http://www.rossmanchance.com/applets/OneSample53.html?population=model to see the
sampling distribution from d.
Solutions:
a. P (800 ≤ X ≤ 900) = normalcdf(800, 900, 845, 35) = 0.8427
b. P (X < 800 or X > 900) = 1 − P (800 ≤ X ≤ 900) = 1 − 0.8427 = 0.1573
c. Q1 = invNorm(0.25, 845, 35) = 821.4 lb
d. The middle 99.7% corresponds to plus or minus 3 standard deviations from the mean of a normal
curve. The sampling distribution of average weight in random samples of 4 bears has mean µ = 845
35
= 17.5, so the range we are looking for is 845−3(17.5) to 845+3(17.5),
and standard deviation σ = √
4
that is 792.5 to 897.5 pounds.
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Lars Hertel
STATS 7
Discussion 5
2
A recent report by the World Health Organization using US Census data states that the distribution of
life expectancy (age at death) in the United States has a mean of 79.8 years and a standard deviation
of 15.5 years. Another study reports that, in a random sample of 15,000 recently deceased US citizens,
the mean age at death is 80.4 years. Which of the following is true?
A) 79.8 is a parameter and 80.4 is a statistic.
B) 79.8 is a statistic and 80.4 is a parameter.
C) Both 79.8 and 80.4 are parameters.
D) Both 79.8 and 80.4 are statistics.
Solution: A
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Cats live for 14 years on average, with a standard deviation of 2 years. A random sample of 60 recently
deceased cats is selected, and the sample mean age at death x̄ of these cats is computed. We know
the random variable x̄ has an approximately normal distribution because of
A) the law of large numbers.
B) the fact that probability is the long-run proportion of times an event occurs.
C) the 68-95-99.7 rule.
D) the central limit theorem.
Solution: D
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Assume we are taking two samples of 100 UCI students each and measure their heights. Which of the
following are different:
A) The standard error of the sample mean in sample 1 and sample 2
B) The population mean of sample 1 and sample 2
C) The standard deviation of the sample mean in sample 1 and sample 2
D) The sample mean in sample 1 and sample 2
Solution: A, D
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Lars Hertel
STATS 7
Discussion 5
5
Assume you buy a bag of candy. The candy has 25 pieces in it. Your bag contains 10 orange pieces,
10 red pieces and 5 yellow pieces.
(a) What is the proportion of orange candy in your bag?
(b) What is the standard error of this sample proportion?
(c) Calculate a 95% confidence interval for the population proportion. Assume that the manufacturer
claims that on average each bag contains 50% orange candy. Would you disagree?
(d) Go to http://www.rossmanchance.com/applets/OneProp/OneProp.htm?candy=1 to see the distribution of orange candy proportions
Solutions:
(a) 10/25√
= 0.4
(b) se = 0.4(1 − 0.4)/25 = 0.098
(c) 0.4 ± 1.960.098 = (0.208, 0.592). There is no evidence that the population proportion is not 50%.
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A CNN/Time poll of 1003 American adults, conducted by telephone in June 2002, was designed to
measure beliefs about apocalyptic predictions. One of the results reported was that 59% of the sample
said that they believe the world will come to an end.
(a) Calculate the standard error of the sample proportion who believe that the world will come to an
end.
(b) Calculate a 95% confidence interval for the population proportion (see CH 10 noteset page 4).
Interpret the interval in a way that could be understood by a statistically naive reader.
Solutions: √
√
(a) s.e.(p̂) = p̂(1 − p̂)/n = .59(1 − .59)/1003 = 0.016
(b)
p̂ ± 1.96 ∗ s.e.(p̂) = 0.59 ± 0.0314 = (0.559, 0.621)
Therefore, we are 95% confident that between 55.9% and 62.1% of all Americans think the world will
come to an end.
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Lars Hertel
STATS 7
Discussion 5
7
[From Practice Midterm A] According to a report about California single-family water use, the daily
household indoor water use, X, can be well modeled using a normal distribution with a mean of 175
gallons per day (gpd) and a standard deviation 8 gpd. For questions requiring a calculator, be sure
to show what you typed in your calculator along with your answers.
a. Calculate P (X = 175).
b. What is the probability that a households daily indoor water use is greater than 180 gpd?
c. How much indoor water is used daily by the 7% highest users?
d. Sketch the sampling distribution of the average daily indoor water use in random samples of 100
California households. Include the values of the mean and standard deviation on your sketch.
Solutions:
a. P (X = 175) = 0 because X is continuous
b. P (X > 180) = normalcdf(180, 1E99, 175, 8) = 0.2660
c. X = invNorm(0.93, 175, 8) = 186.8, so the 7% highest users consume about 187 gpd and more
d.
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