AP Statistics Sampling Distributions — Worked Answer Explanations

Unit 5 · 12 questions explained

1. Question 1 · Easy

   A population has mean $\mu =50$ and standard deviation $\sigma =10$. A simple random sample of size $n=25$ is taken. What is the standard deviation of the sampling distribution of $\bar{x}$?

   • A
     10

     Why not A: This is $\sigma$, the population standard deviation, not the standard error of $\bar{x}$.
   • B
     2Correct
   • C
     0.4

     Why not C: Dividing $\sigma$ by $n=25$ instead of $\sqrt{n}=5$.
   • D
     50

     Why not D: This is the population mean, not the standard error.
   Explanation

   The standard error (standard deviation of the sampling distribution of $\bar{x}$) is ${\sigma }_{\bar{x}}=\sigma /\sqrt{n}=10/\sqrt{25}=10/5=2$.

   Key takeaway

   Standard error of $\bar{x}$: $\sigma_{\bar{x}} = \sigma/\sqrt{n}$. Larger samples reduce the standard error.

2. Question 2 · Easy

   Which statement best describes what the Central Limit Theorem (CLT) tells us about the sampling distribution of $\bar{x}$?

   • A
     The population must be normally distributed for $\bar{x}$ to be approximately normal.

     Why not A: The CLT applies to non-normal populations; the entire point is that normality of the population is not required for large $n$.
   • B
     For large enough sample sizes, the sampling distribution of $\bar{x}$ is approximately normal regardless of the population shape.Correct
   • C
     For large samples, the sample mean equals the population mean.

     Why not C: $E(\bar{x})=\mu$ for any sample size; the CLT specifically says the sampling distribution is approximately normal for large $n$.
   • D
     As sample size increases, $\bar{x}$ converges to $\sigma$.

     Why not D: $\bar{x}$ converges to $\mu$ (the Law of Large Numbers); it has no relationship to $\sigma$.
   Explanation

   The Central Limit Theorem states that for sufficiently large $n$ (often $n\ge 30$ as a rule of thumb), the sampling distribution of $\bar{x}$ is approximately normal with mean $\mu$ and standard error $\sigma /\sqrt{n}$, regardless of the shape of the population distribution.

   Key takeaway

   CLT: for large $n$, $\bar{x} \sim N(\mu,\, \sigma/\sqrt{n})$ approximately, even if the population is not normal.

3. Question 3 · Easy

   A large population has proportion $p=0.60$ with a certain characteristic. A random sample of $n=100$ is taken. What is the mean of the sampling distribution of $\hat{p}$?

   • A
     0.060

     Why not A: Dividing $p$ by $n$ gives a standard error-like value, not the mean.
   • B
     60

     Why not B: This is the expected count $np$, not the mean of the sampling distribution of $\hat{p}$.
   • C
     0.60Correct
   • D
     0.049

     Why not D: 0.049 is the standard deviation of $\hat{p}$, not its mean.
   Explanation

   The sampling distribution of $\hat{p}$ is centered at $p$, the population proportion. So ${\mu }_{\hat{p}}=p=0.60$. The sample proportion is an unbiased estimator of the population proportion.

   Key takeaway

   The mean of the sampling distribution of $\hat{p}$ is $p$; $\hat{p}$ is an unbiased estimator of $p$.

4. Question 4 · Easy

   A population of SAT scores is strongly right-skewed. Samples of size $n=5$ are taken repeatedly. Which of the following best describes the sampling distribution of $\bar{x}$ for $n=5$?

   • A
     Approximately normal, because the CLT applies for any sample size.

     Why not A: The CLT requires sufficiently large $n$ (approximately 30+) for a skewed population; $n=5$ is too small.
   • B
     Exactly normal, because all sampling distributions of $\bar{x}$ are normal.

     Why not B: Sampling distributions of $\bar{x}$ are exactly normal only when the population is normal.
   • C
     Still right-skewed, but less skewed than the population.Correct
   • D
     Identical in shape to the population distribution.

     Why not D: Averaging reduces skewness even for small samples; the sampling distribution is never identical to the population shape when $n>1$.
   Explanation

   For small samples from a skewed population, the CLT has not yet "kicked in." The sampling distribution of $\bar{x}$ will be less skewed than the population (averaging reduces extremes), but with $n=5$ it will still be noticeably right-skewed.

   Key takeaway

   The CLT requires larger samples for skewed populations; $n \geq 30$ is a common threshold, though the required $n$ increases with skewness.

5. Question 5 · Easy

   A population proportion is $p=0.40$. A random sample of $n=100$ is drawn. What is the standard deviation of the sampling distribution of $\hat{p}$?

   • A
     0.40

     Why not A: This is $p$ itself, not the standard deviation of the sampling distribution.
   • B
     0.0024

     Why not B: Computing $p(1-p)/n=0.0024$ gives the variance, not the standard deviation.
   • C
     0.049Correct
   • D
     0.490

     Why not D: Forgetting to divide by $\sqrt{n}$: $\sqrt{0.40\times 0.60}=0.490$ is just $\sqrt{pq}$.
   Explanation

   ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40\times 0.60}{100}}=\sqrt{\frac{0.24}{100}}=\sqrt{0.0024}\approx 0.049$.

   Key takeaway

   Standard deviation of $\hat{p}$: $\sigma_{\hat{p}} = \sqrt{p(1-p)/n}$.

6. Question 6 · Easy

   A researcher takes random samples of size $n=64$ from a population with $\mu =80$ and $\sigma =16$. What is the probability that a sample mean exceeds 82?

   • A
     $P(Z>2.0)\approx 0.023$

     Why not A: This uses $\sigma =16$ without dividing by $\sqrt{n}$: $z=(82-80)/16=0.125$, not 2.
   • B
     $P(Z>1.0)\approx 0.159$Correct
   • C
     $P(Z>0.125)\approx 0.450$

     Why not C: Using $\sigma$ directly instead of the standard error $\sigma /\sqrt{n}=2$.
   • D
     $P(Z>4.0)\approx 0.00003$

     Why not D: Dividing by $n=64$ instead of $\sqrt{n}=8$, giving $z=(82-80)/(16/64)=8$, not 4.
   Explanation

   Standard error: ${\sigma }_{\bar{x}}=16/\sqrt{64}=2$. $z=(82-80)/2=1.0$. $P(\bar{x}>82)=P(Z>1.0)\approx 0.159$.

   Key takeaway

   To find probabilities for $\bar{x}$, standardize using $z = (\bar{x} - \mu)/(\sigma/\sqrt{n})$.

7. Question 7 · Easy

   Which condition must be met for the sampling distribution of $\hat{p}$ to be approximately normal?

   • A
     $n\ge 30$

     Why not A: $n\ge 30$ is the CLT rule of thumb for the sampling distribution of $\bar{x}$, not $\hat{p}$.
   • B
     $np\ge 10$ and $n(1-p)\ge 10$Correct
   • C
     $p\ge 0.5$

     Why not C: There is no requirement that $p$ be close to 0.5; the large-counts condition involves $np$ and $n(1-p)$.
   • D
     The population must be normally distributed.

     Why not D: Proportions come from a binary categorical variable; population normality is irrelevant.
   Explanation

   The sampling distribution of $\hat{p}$ is approximately normal when the Large Counts condition is met: $np\ge 10$ and $n(1-p)\ge 10$. This ensures enough expected successes and failures for the normal approximation to be valid.

   Key takeaway

   Large Counts condition: $np \geq 10$ and $n(1-p) \geq 10$ for $\hat{p}$ to be approximately normal.

8. Question 8 · Easy

   A population is normal with $\mu =100$ and $\sigma =15$. Random samples of size $n=9$ are repeatedly drawn. What is the shape of the sampling distribution of $\bar{x}$?

   • A
     Approximately normal, by the CLT since $n=9\ge 5$.

     Why not A: The CLT is not needed here; exact normality follows from the population being normal, for any $n$.
   • B
     Exactly normal, because the population is normal.Correct
   • C
     Unknown shape, since $n<30$.

     Why not C: The $n\ge 30$ rule applies to non-normal populations; when the population is normal, $\bar{x}$ is exactly normal for any $n$.
   • D
     Approximately normal with mean 100/9.

     Why not D: The mean of $\bar{x}$ is $\mu =100$, not $\mu /n$.
   Explanation

   If the population is normally distributed, the sampling distribution of $\bar{x}$ is exactly normal for any sample size $n$. The CLT (which gives approximate normality) is only needed when the population is not normal.

   Key takeaway

   If the population is normal, $\bar{x}$ is exactly normal for any $n$; the CLT provides only approximate normality for non-normal populations with large $n$.

9. Question 9 · Medium

   A sample of $n=50$ is taken from a large population where $p=0.20$. A student checks the conditions for using a normal approximation and finds $np=10$, $n(1-p)=40$. The student concludes the approximation is barely acceptable. Is this correct?

   • A
     No; the condition requires $np>10$ strictly, so $np=10$ fails.

     Why not A: The large-counts condition uses $\ge 10$, not $>10$ strictly.
   • B
     Yes; both $np\ge 10$ and $n(1-p)\ge 10$ are satisfied, so the normal approximation is justified.Correct
   • C
     No; the condition also requires $n\ge 30$, which fails here.

     Why not C: The $n\ge 30$ condition applies to the mean $\bar{x}$, not to $\hat{p}$.
   • D
     No; for a proportion, the sample size must be at least 100.

     Why not D: There is no fixed minimum sample size of 100 for proportions; the large-counts condition depends on $p$ and $n$ together.
   Explanation

   The standard Large Counts condition is $np\ge 10$ AND $n(1-p)\ge 10$. Here $np=10\ge 10$ and $n(1-p)=40\ge 10$, so both conditions are met. The normal approximation is valid. The student is essentially correct.

   Key takeaway

   Large Counts uses $\geq 10$ (not strict $> 10$); when $np = 10$ exactly, the condition is satisfied.

10. Question 10 · Medium

    Two sampling distributions are described: (I) $n=25$ samples from a symmetric, bell-shaped population; (II) $n=100$ samples from a heavily skewed population. Which is more likely to be approximately normal?

    • A
      Distribution I, because the population is symmetric.

      Why not A: Symmetry helps, but with $n=25$ and bell-shape, the sampling distribution is already very close to normal — the comparison depends on severity of skewness in II.
    • B
      Distribution II, because $n=100$ is large enough by the CLT even for skewed populations.Correct
    • C
      Both are equally normal, because the CLT applies in both cases.

      Why not C: The CLT does not apply equally; $n=25$ from a near-normal population and $n=100$ from a skewed population may differ, but for heavily skewed data $n=100$ typically suffices.
    • D
      Neither is approximately normal, because neither comes from a perfectly normal population.

      Why not D: The CLT gives approximate normality for large $n$; exact population normality is not required.
    Explanation

    For a symmetric, bell-shaped population with $n=25$, the sampling distribution is already nearly normal. For a heavily skewed population, $n=100$ generally satisfies the CLT condition (skewed distributions often require $n\ge 30$ to $\ge 50$+). Both are approximately normal, but the comparison question is asking which is more likely — with heavy skew and $n=100$, normality is well established.

    Key takeaway

    For heavily skewed populations, larger samples (often $n \geq 30–50$) are needed for the CLT to produce an approximately normal sampling distribution of $\bar{x}$.

11. Question 11 · Hard

    A factory produces bolts with a mean diameter of $\mu =10$ mm and $\sigma =0.5$ mm. Quality control selects random samples of 16 bolts. In repeated sampling, 95% of sample means will fall within what range?

    • A
      9.00 mm to 11.00 mm

      Why not A: Using $\pm 2\sigma$ instead of $\pm 2\sigma /\sqrt{n}$.
    • B
      9.75 mm to 10.25 mmCorrect
    • C
      9.50 mm to 10.50 mm

      Why not C: Using $\pm 2(0.5)/\sqrt{4}=\pm 0.5$ — dividing by $\sqrt{4}$ instead of $\sqrt{16}$.
    • D
      9.875 mm to 10.125 mm

      Why not D: Using $z^{\ast }=1.0$ instead of $z^{\ast }=1.96\approx 2$ for 95%.
    Explanation

    Standard error $=\sigma /\sqrt{n}=0.5/\sqrt{16}=0.5/4=0.125$ mm. For 95%, use $z^{\ast }\approx 2$: interval $=10\pm 2(0.125)=10\pm 0.25$, giving $9.75$ to $10.25$ mm.

    Key takeaway

    95% of sample means fall within $\mu \pm 2(\sigma/\sqrt{n})$; use the standard error, not $\sigma$.

12. Question 12 · Hard

    A sample proportion $\hat{p}=0.55$ is calculated from a sample of $n=200$. The population proportion is claimed to be $p=0.50$. What is the approximate $z$-score for $\hat{p}$?

    • A
      $z\approx 1.41$Correct
    • B
      $z\approx 2.00$

      Why not B: Using ${\sigma }_{\hat{p}}=\sqrt{\hat{p}(1-\hat{p})/n}$ with $\hat{p}=0.55$ gives $\sigma \approx 0.035$, not 0.025.
    • C
      $z\approx 0.14$

      Why not C: Not dividing the numerator by the standard error: $(0.55-0.50)=0.05$ is the raw difference.
    • D
      $z\approx 3.16$

      Why not D: Using $\sqrt{n}$ directly in numerator: $0.05\times \sqrt{200}\approx 0.71$; this is not the correct formula.
    Explanation

    Under $H_0:p=0.50$, the standard deviation of $\hat{p}$ is ${\sigma }_{\hat{p}}=\sqrt{p(1-p)/n}=\sqrt{0.50\times 0.50/200}=\sqrt{0.00125}\approx 0.0354$. $z=(\hat{p}-p)/{\sigma }_{\hat{p}}=(0.55-0.50)/0.0354\approx 1.41$.

    Key takeaway

    For a hypothesis test on $p$, use $\sigma_{\hat{p}} = \sqrt{p_0(1-p_0)/n}$ (the null value $p_0$, not $\hat{p}$).