Using the Central Limit Theorem Quiz
The question sheet
Reveal any answer as you study-
If you are asked to find the probability of an individual value, you should:
- Use the CLT for the mean
- Not use the CLT; use its distribution
- Use the CLT for sums
- Use the law of large numbers
Reveal answer
Answer: Not use the CLT; use its distribution
Source evidence
PDF page 416: It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.
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When asked to find the probability of a sum or total, which method do you use?
- CLT for sums
- Distribution of the individual
- CLT for the mean
- Normal approximation to binomial
Reveal answer
Answer: CLT for sums
Source evidence
PDF page 416: It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.
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The law of large numbers says that as sample size increases, the sample mean tends to:
- Get closer to μ
- Grow without bound
- Stay constant
- Move away from μ
Reveal answer
Answer: Get closer to μ
Source evidence
PDF page 417: Law of Large Numbers ̄ The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x of the sample tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the ̄ ̄ σ standard deviation for X is .) This means that the sample mean x must be close to the population mean μ. We can n say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. Central Limit Theorem for the Mean and Sum Examples
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As n gets larger, the standard deviation of the sample means:
- Gets larger
- Gets smaller
- Stays the same
- Becomes negative
Reveal answer
Answer: Gets smaller
Source evidence
PDF page 417: Law of Large Numbers ̄ The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x of the sample tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the ̄ ̄ σ standard deviation for X is .) This means that the sample mean x must be close to the population mean μ. We can n say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. Central Limit Theorem for the Mean and Sum Examples
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For the stress study, the individual stress score follows which distribution?
- X ~ Exp(5)
- X ~ U(1, 5)
- X ~ N(1, 5)
- X ~ B(1, 5)
Reveal answer
Answer: X ~ U(1, 5)
Source evidence
PDF page 417: Let X = one stress score. Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution). a + b 1 + 5 = = 3
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For the stress example, the mean of a single stress score equals:
- 1.15
- 1
- 5
- 3
Reveal answer
Answer: 3
Source evidence
PDF page 417: Let X = one stress score. Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution). a + b 1 + 5 = = 3
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In the stress study, what is the probability the mean score of 75 students is less than two?
- About zero
- About 0.5
- About one
- About 0.9
Reveal answer
Answer: About zero
Source evidence
PDF page 417: The probability that the mean stress score is less than two is about zero.
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The 90th percentile for the mean of 75 stress scores is about:
- 3.2
- 5.0
- 3.0
- 2.0
Reveal answer
Answer: 3.2
Source evidence
PDF page 418: The 90 percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2. ⎛ 1.15⎞ invNorm 0.90,3, = 3.2 ⎝ 75 ⎠ For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3), ( 75) (1.15)]
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The mean of the sum of 75 stress scores equals:
- 75
- 225
- 237.8
- 200
Reveal answer
Answer: 225
Source evidence
PDF page 418: c. The mean of the sum of 75 stress scores is (75)(3) = 225
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The standard deviation of the sum of 75 stress scores is about:
- 9.96
- 225
- 1.15
- 3.0
Reveal answer
Answer: 9.96
Source evidence
PDF page 419: The standard deviation of the sum of 75 stress scores is ( 75) (1.15) = 9.96 P(Σx < 200) = 0
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For the cell phone study, the excess time used by one customer follows which distribution?
- Normal with mean 80
- Uniform between 0 and 22
- Exponential with mean 22
- Binomial with n = 80
Reveal answer
Answer: Exponential with mean 22
Source evidence
PDF page 420: Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. ⎛ 1 ⎞ X ∼ Exp
PDF page 420: . From previous chapters, we know that μ = 22 and σ = 22.
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For 80 customers, the probability the mean excess time is more than 20 minutes is about:
- 0.4029
- 0.95
- 0.7919
- 0.0741
Reveal answer
Answer: 0.7919
Source evidence
PDF page 420: ⎛ ̄ 22 ⎞ P( x > 20) = 0.79199 using normalcdf 20,1E99,22, ⎝ 80⎠ The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
Introductory Statistics
Introductory Statistics by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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