The Central Limit Theorem for Sample Means (Averages) Quiz
The question sheet
Reveal any answer as you study-
According to the central limit theorem, as sample size n increases, the distribution of sample means tends toward what?
- An exponential distribution
- A normal distribution
- A uniform distribution
- A skewed distribution
Reveal answer
Answer: A normal distribution
Source evidence
PDF page 408: ̄ If you draw random samples of size n, then as n increases, the random variable X which consists of sample means, tends
PDF page 409: ⎛ σx⎞ . X ~ N μ x , ⎝ n⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ̄ To put it more formally, if you draw random samples of size n, the distribution of the random variable X , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. ̄ ̄ The random variable X has a different z-score associated with it from that of the random variable X. The mean x is the ̄ value of X in one sample. ̄ x − μ x z =
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In the central limit theorem, the sampling distribution of the mean has what mean compared to the original distribution?
- Half the mean
- The same mean
- A larger mean
- A smaller mean
Reveal answer
Answer: The same mean
Source evidence
PDF page 409: ⎛ σx⎞ . X ~ N μ x , ⎝ n⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ̄ To put it more formally, if you draw random samples of size n, the distribution of the random variable X , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. ̄ ̄ The random variable X has a different z-score associated with it from that of the random variable X. The mean x is the ̄ value of X in one sample. ̄ x − μ x z =
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The variance of the sampling distribution equals the original variance divided by what?
- n squared
- The square root of n
- The sample size n
- Two times n
Reveal answer
Answer: The sample size n
Source evidence
PDF page 409: ⎛ σx⎞ . X ~ N μ x , ⎝ n⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ̄ To put it more formally, if you draw random samples of size n, the distribution of the random variable X , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. ̄ ̄ The random variable X has a different z-score associated with it from that of the random variable X. The mean x is the ̄ value of X in one sample. ̄ x − μ x z =
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The standard deviation of the sampling distribution equals the original standard deviation divided by what?
- n squared
- The square root of n
- The sample size n
- The variance
Reveal answer
Answer: The square root of n
Source evidence
PDF page 409: ⎛ σx⎞ . X ~ N μ x , ⎝ n⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ̄ To put it more formally, if you draw random samples of size n, the distribution of the random variable X , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. ̄ ̄ The random variable X has a different z-score associated with it from that of the random variable X. The mean x is the ̄ value of X in one sample. ̄ x − μ x z =
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In the CLT, the variable n represents:
- The population mean
- The number of experiments done
- The number of values averaged together
- The original standard deviation
Reveal answer
Answer: The number of values averaged together
Source evidence
PDF page 409: ⎛ σx⎞ . X ~ N μ x , ⎝ n⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ̄ To put it more formally, if you draw random samples of size n, the distribution of the random variable X , which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. ̄ ̄ The random variable X has a different z-score associated with it from that of the random variable X. The mean x is the ̄ value of X in one sample. ̄ x − μ x z =
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The standard deviation of the sample mean, σx/√n, is also called the:
- Sampling ratio
- Decay parameter
- Population variance
- Standard error of the mean
Reveal answer
Answer: Standard error of the mean
Source evidence
PDF page 409: = standard deviation of X and is called the standard error of the mean. n
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With μX=90, σX=15, and n=25, the probability the sample mean is between 85 and 92 is:
- 0.5000
- 0.6997
- 0.9977
- 0.9962
Reveal answer
Answer: 0.6997
Source evidence
PDF page 410: ̄ ⎛ 15 ⎞ X ~ N 90, . ⎝ 25⎠ ̄ Find P(85 < x < 92). Draw a graph. ̄ P(85 < x < 92) = 0.6997 The probability that the sample mean is between 85 and 92 is 0.6997.
PDF page 410: normalcdf(lower value, upper value, mean, standard error of the mean) σ The parameter list is abbreviated (lower value, upper value, μ, ) n 15 normalcdf(85,92,90, ) = 0.6997 25
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For μX=90, σX=15, n=25, the value two standard deviations above the expected mean 90 is:
- 99
- 105
- 93
- 96
Reveal answer
Answer: 96
Source evidence
PDF page 410: ⎝ n⎠ ⎛ 15 ⎞ value = 90 + 2 = 96 ⎝ 25⎠ The value that is two standard deviations above the expected value is 96. 15 σx = = 3. Recall that the standard error of the mean is a description of The standard error of the mean is n 25 how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n.
PDF page 410: value = μx + (#ofTSDEVs)
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For the sample with σX=15 and n=25, the standard error of the mean equals:
- 5
- 15
- 0.6
- 3
Reveal answer
Answer: 3
Source evidence
PDF page 410: ⎝ n⎠ ⎛ 15 ⎞ value = 90 + 2 = 96 ⎝ 25⎠ The value that is two standard deviations above the expected value is 96. 15 σx = = 3. Recall that the standard error of the mean is a description of The standard error of the mean is n 25 how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n.
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In normalcdf, the 'mean' parameter refers to the mean of:
- The sampling distribution
- The z-score table
- The sample data only
- The original distribution
Reveal answer
Answer: The original distribution
Source evidence
PDF page 409: • mean is the mean of the original distribution
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For the soccer match example with μX=2, σX=0.5, n=50, P(1.8<x<2.3) equals:
- 0.9962
- 0.9977
- 0.5000
- 0.6997
Reveal answer
Answer: 0.9977
Source evidence
PDF page 411: μX = 2, σX = 0.5, n = 50, and X ~ N 2,
PDF page 411: ⎝ 50⎠ ̄ Find P(1.8 < x < 2.3). Draw a graph. ̄ P(1.8 < x < 2.3) = 0.9977 ⎛ .5 ⎞ normalcdf 1.8,2.3,2, = 0.9977 ⎝ 50⎠ The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.
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To find percentiles for sample means on the calculator, which function is used?
- poissoncdf
- binomcdf
- normalcdf
- invNorm
Reveal answer
Answer: invNorm
Source evidence
PDF page 411: standard deviation⎞ k = invNorm area to the left of k, mean,
PDF page 411: To find percentiles for means on the calculator, follow these steps.
Introductory Statistics
Introductory Statistics by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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