Independent rolls of a fair six-sided die have mean μ = 3.5. The first three results are 2, 4, 3, and the fourth is 1. LLN says adding the fourth result cannot move the running mean farther from μ. Is this true?

Circle one:   True   False

Independent observations from one population have standard deviation 16. Sample size rises from 4 to 64. Find the new standard error divided by the old standard error. Give your answer as a decimal.

Answer: ______________

Take 49 independent observations from one population with standard deviation 14. What is the standard error of their sample mean?

Answer: ______________

A strongly skewed population has finite, positive variance. A researcher takes 100 independent observations. Does this alone certify an accurate normal approximation for the mean?

• Yes; a sample size of 100 guarantees accuracy for every population with finite variance.
• Yes; independent observations make the mean exactly normal regardless of population shape.
• No; convergence is a limit as n grows, so accuracy at n = 100 still depends on population shape.

Nine independent observations come from a normal population with mean 60 and standard deviation 15. The sample mean is 70. Find its standardised value z.

Answer: ______________

Independent observations come from one population with mean μ and standard deviation σ > 0. A normal model fits the sample mean. How do you standardise it?

• Subtract σ from the sample mean, then divide by the population mean μ.
• Subtract μ from the sample mean, then divide by its standard error σ / √n.
• Subtract μ from the sample mean, then divide by the population standard deviation σ.

Independent observations from one population have standard deviation 10. What is the smallest integer sample size giving a standard error of the mean at most 3?

Answer: ______________

A random observation X has standard deviation 7. A researcher fills all 100 entries of a sample with copies of this same X. What is the standard deviation of their mean?

• 0.7, using 7 / √100 as if the entries were independent.
• 0, because equal entries make the mean nonrandom.
• 7, since averaging these dependent copies leaves the random observation X unchanged.
• 0.07, using 7 / 100 as the spread of the mean of 100 entries.

An income population is skewed and has finite, positive variance. Dee says the CLT makes one randomly selected income normally distributed. Which response is valid?

• One income becomes normal if the population has finite, positive variance.
• One income becomes normal when many other incomes are sampled independently.
• One income keeps the population distribution; CLT concerns standardised sample means.

For 200 independent trials with success probability p = 0.03 on every trial, does the check requiring both expected counts to be at least 10 support a normal model for the sample proportion?

• No: 6 expected successes fall below the required minimum of 10.
• Yes: a total of 200 trials is enough.
• Yes: 194 expected failures satisfy the check regardless of expected successes.