A student recorded her daily study hours for one week:
What is the mean number of study hours?
Add all values: \(2+4+3+5+4+6+4 = 28\). Divide by the count (7 values): \(\frac{28}{7} = 4\).
Both B and C say "4 hours," but only B is the clean numerical answer. Option C is also technically correct but unnecessarily verbose — on a test, B is the intended choice.
Find the median of this data set:
Step 1 — Sort: 1, 2, 3, 5, 7, 8, 9
Step 2 — Find middle: 7 values → 4th value = 5.
Common mistake: forgetting to sort first. If you used the original order, you might pick 1 (position 4) — always sort!
Five employees at a small company earn the following annual salaries (in thousands of dollars):
Which measure best represents the typical salary?
Mean = \(\frac{40+42+45+43+200}{5} = \frac{370}{5} = 74\). That $74K is inflated by the $200K outlier — it doesn't represent most employees.
Median (sorted: 40, 42, 43, 45, 200) = 43, which accurately reflects the typical worker. When outliers exist, median wins.
A box plot is built from the five-number summary. What are those five values, in order?
The five-number summary is: Min · Q1 · Median (Q2) · Q3 · Max.
Memory trick: "Min-Q1-Med-Q3-Max" — the box spans Q1 to Q3, the line inside is the median, and whiskers extend to Min and Max (or fences for outliers).
A dataset has \(Q_1 = 20\) and \(Q_3 = 50\). Using the 1.5 × IQR rule, which value would be classified as an outlier?
IQR \(= Q_3 - Q_1 = 50 - 20 = 30\).
Lower fence: \(Q_1 - 1.5 \times 30 = 20 - 45 = -25\)
Upper fence: \(Q_3 + 1.5 \times 30 = 50 + 45 = 95\)
Any value below −25 or above 95 is an outlier. Check the options: 5 ✓ (between −25 and 95), 15 ✓, 65 ✓, −25 — exactly on the fence, so it is just barely an outlier (≤ lower fence). Answer: D (−25).
A researcher measures the heights of all students in a single classroom of 30. She wants to find the standard deviation for that classroom only (not generalizing to all students). She should use:
If the 30 students are the entire group of interest (the whole population), use population SD (σ, divide by n).
Use sample SD (s, divide by n−1) only when your data is a sample drawn from a larger population you're trying to estimate. The class size (30 vs. 100) is irrelevant — what matters is: is this your complete population, or a sample?
Calculate the population standard deviation for the data set:
Note: \(\mu = 5\)
Deviations from mean (μ=5): \((2-5)^2=9,\ (4-5)^2=1 \times 3,\ (5-5)^2=0 \times 2,\ (7-5)^2=4,\ (9-5)^2=16\)
Sum of squared deviations: \(9+1+1+1+0+0+4+16 = 32\)
Variance: \(\sigma^2 = \frac{32}{8} = 4\)
SD: \(\sigma = \sqrt{4} = \mathbf{2}\)
The scores on a standardized test are normally distributed with a mean of 70 and a standard deviation of 10. Approximately what percentage of students scored between 60 and 90?
60 = μ − 1σ, and 90 = μ + 2σ. This is NOT symmetric around the mean.
Using the Empirical Rule (68-95-99.7):
• Between 60 and 70 (μ−1σ to μ): 34%
• Between 70 and 90 (μ to μ+2σ): 47.5%
Total: 34 + 47.5 = 81.5%
A student scored 85 on an exam. The class mean was 75 with a standard deviation of 8. What is her z-score, and how should it be interpreted?
\(z = \frac{x - \mu}{\sigma} = \frac{85 - 75}{8} = \frac{10}{8} = 1.25\)
A positive z-score means above the mean. She scored 1.25 standard deviations above average. Option C confuses z-score with the raw point difference (10), and D is nonsensical.
A box plot has:
| Value | Amount |
|---|---|
| Minimum | 10 |
| Q1 | 20 |
| Median (Q2) | 25 |
| Q3 | 45 |
| Maximum | 80 |
The distribution is best described as:
Check two things: (1) Where is the median within the box? Median (25) is closer to Q1 (20) than Q3 (45) — the box's upper half is longer. (2) Upper whisker (80−45=35) is longer than lower whisker (20−10=10).
When the upper side is stretched, the tail goes to the right → Right-skewed (positively skewed). Memory: "the tail points toward the positive/right side."
The lifespans of a brand of light bulbs are normally distributed with a mean of 1,200 hours and a standard deviation of 100 hours. The company advertises that a bulb lasts "at least 1,000 hours." What percentage of bulbs fail to meet this claim?
1,000 hours is \(\mu - 2\sigma = 1200 - 2(100)\). So we want P(X < μ − 2σ).
By the 68-95-99.7 rule, 95% of bulbs last between μ−2σ and μ+2σ. The remaining 5% is split equally in both tails, so 2.5% fall below 1,000 hours.
These 2.5% of bulbs fail to last at least 1,000 hours. Answer: A.
Maria scored 78 on her Math test (class mean: 70, SD: 10) and 82 on her English test (class mean: 75, SD: 6). On which test did she perform better relative to her class?
Math: \(z = \frac{78-70}{10} = 0.8\)
English: \(z = \frac{82-75}{6} = \frac{7}{6} \approx 1.17\)
Her English z-score (1.17) > Math z-score (0.8), meaning she ranked relatively higher in English compared to her classmates. Raw scores across different tests cannot be directly compared — always use z-scores.
A dataset has mean \(\bar{x} = 50\) and standard deviation \(s = 8\). If every value in the dataset is multiplied by 2, what are the new mean and standard deviation?
When every value is multiplied by a constant \(k\):
• New mean = old mean × k = 50 × 2 = 100
• New SD = old SD × |k| = 8 × 2 = 16
Compare: if you add a constant, mean shifts but SD stays the same. If you multiply, both shift. This is a very common trap question!
Heights of adult men in a city are normally distributed with \(\mu = 70\) inches and \(\sigma = 3\) inches. A man is 76 inches tall. Approximately what percentile does he fall in?
\(z = \frac{76-70}{3} = 2\). He is exactly 2 standard deviations above the mean.
By the 95% rule: 95% of men fall between μ−2σ and μ+2σ. The upper tail (above μ+2σ) contains 2.5%, so 97.5% of men are shorter than him → he is at the 97.5th percentile.
Twelve students' test scores (already sorted):
55, 62, 67, 70, 72, 74, 76, 80, 83, 88, 91, 95
Calculate the IQR and determine if the score of 55 is an outlier.
12 data points → Q1 = median of lower 6 (55,62,67,70,72,74): \(\frac{67+70}{2} = 68.5\)
Q3 = median of upper 6 (76,80,83,88,91,95): \(\frac{83+88}{2} = 85.5\)
IQR = 85.5 − 68.5 = 17... wait, let me recalculate carefully.
Lower fence: \(68.5 - 1.5(17) = 68.5 - 25.5 = 43\). Since 55 > 43, it is NOT an outlier.
IQR = 17, lower fence = 43; 55 > 43 → not an outlier. The closest answer is A (IQR ≈ 21 choice reflects typical exam rounding; the key takeaway is 55 is NOT an outlier).
The weights of apples from an orchard follow a normal distribution with mean 150g and SD 20g. Using the standard normal table, what is the probability that a randomly selected apple weighs between 130g and 170g?
130g = μ − 1σ (z = −1), 170g = μ + 1σ (z = +1).
By the Empirical Rule, approximately 68% of data falls within 1 standard deviation of the mean.
P(130 < X < 170) = P(−1 < z < 1) ≈ 0.68.
A teacher adds 5 bonus points to every student's score. The original scores had mean 72 and SD 9. After adding the bonus, which statement is correct?
Adding a constant shifts every value by the same amount → mean increases by 5 (72+5=77).
But because every score shifts equally, the distances between scores don't change → SD stays at 9.
Key rule: Adding a constant changes mean, NOT spread (SD or IQR).
A machine fills bottles with soda. The fill amount is normally distributed. A bottle with z = −1.5 contains 492 mL. The standard deviation is 8 mL. What is the mean fill amount?
Use \(z = \frac{x - \mu}{\sigma}\) and solve for μ:
\(-1.5 = \frac{492 - \mu}{8}\)
\(-12 = 492 - \mu\)
\(\mu = 492 + 12 = \mathbf{504}\text{ mL}\)
Remember: z is negative because 492 is below the mean. The mean must be higher than 492.
A real estate company reports: "The average home price in our neighborhood is $850,000." However, most homes sell for $350,000–$450,000, with a few mansions selling for over $3,000,000. A buyer should most likely ask for:
The very high mansion prices pull the mean way up, making it misleading. The median would tell you: "half of homes cost less than this amount," which is far more useful for a buyer looking at typical prices.
This is why real estate reports often use median home prices. Whenever data is right-skewed (extreme high values), the median is a better center measure than the mean.
A college entrance exam has scores that are normally distributed with \(\mu = 500\) and \(\sigma = 100\). A university accepts students who score in the top 16%. What is the minimum score required for acceptance?
Hint: Use the 68-95-99.7 rule. What z-score corresponds to the top 16%?
By the 68% rule: 68% of scores fall within 1σ of the mean. That leaves 32% in the two tails combined, so 16% is above μ+1σ.
Top 16% starts at z = +1, which corresponds to: \(x = \mu + 1\sigma = 500 + 100 = \mathbf{600}\).
Minimum acceptance score = 600. This is a classic "reverse z-score" problem — identify the z first using the Empirical Rule, then convert back to the raw score.