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01
A student scored 72, 85, 90, 68, and 95 on five math tests.
What is the mean of these scores?
What is the mean of these scores?
✓ Correct! The answer is B — 82.
Add all scores: 72 + 85 + 90 + 68 + 95 = 410. Divide by 5: 410 ÷ 5 = 82.
Memory: Mean = (Sum of all values) ÷ (number of values)
Add all scores: 72 + 85 + 90 + 68 + 95 = 410. Divide by 5: 410 ÷ 5 = 82.
Memory: Mean = (Sum of all values) ÷ (number of values)
02
Find the median of the following dataset:
3, 7, 1, 9, 4, 6, 2
3, 7, 1, 9, 4, 6, 2
Hint: Remember to arrange values in order first.
✓ The answer is C — 4.
Sorted order: 1, 2, 3, 4, 6, 7, 9. With 7 values, the median is the 4th value = 4.
Memory: Sort first, pick the MIDDLE value. For odd count: middle position = (n+1)/2.
Sorted order: 1, 2, 3, 4, 6, 7, 9. With 7 values, the median is the 4th value = 4.
Memory: Sort first, pick the MIDDLE value. For odd count: middle position = (n+1)/2.
03
A dataset has Q1 = 20 and Q3 = 45.
What is the Interquartile Range (IQR)?
What is the Interquartile Range (IQR)?
✓ The answer is B — 25.
IQR = Q3 − Q1 = 45 − 20 = 25.
Memory: IQR = Q3 − Q1. It measures the spread of the MIDDLE 50% of data.
IQR = Q3 − Q1 = 45 − 20 = 25.
Memory: IQR = Q3 − Q1. It measures the spread of the MIDDLE 50% of data.
04
A dataset has Q1 = 10, Q3 = 30, and IQR = 20.
Which of the following values would be considered an outlier?
Which of the following values would be considered an outlier?
Use the 1.5 × IQR rule to find the fences.
✓ The answer is D — 62.
Upper fence = Q3 + 1.5 × IQR = 30 + 1.5(20) = 30 + 30 = 60.
Lower fence = Q1 − 1.5 × IQR = 10 − 30 = −20.
Only 62 exceeds the upper fence of 60 → outlier!
Note: 55 < 60, so 55 is NOT an outlier. 5 > −20, so 5 is NOT an outlier.
Upper fence = Q3 + 1.5 × IQR = 30 + 1.5(20) = 30 + 30 = 60.
Lower fence = Q1 − 1.5 × IQR = 10 − 30 = −20.
Only 62 exceeds the upper fence of 60 → outlier!
Note: 55 < 60, so 55 is NOT an outlier. 5 > −20, so 5 is NOT an outlier.
05
A real estate agent records home prices in a neighborhood:
$180k, $190k, $195k, $200k, $205k, $210k, $850k
Which statement is most accurate?
$180k, $190k, $195k, $200k, $205k, $210k, $850k
Which statement is most accurate?
✓ The answer is C.
Mean ≈ (180+190+195+200+205+210+850)/7 ≈ $290k. The $850k outlier pulls the mean up drastically.
Median = 4th value (sorted) = $200k.
The median of $200k better represents a "typical" home price. With outliers, use median, not mean.
Mean ≈ (180+190+195+200+205+210+850)/7 ≈ $290k. The $850k outlier pulls the mean up drastically.
Median = 4th value (sorted) = $200k.
The median of $200k better represents a "typical" home price. With outliers, use median, not mean.
06
A researcher surveys 25 students from a school of 600 to estimate the average hours of sleep per night. She calculates a standard deviation from this data.
Which formula and symbol should she use?
Which formula and symbol should she use?
✓ The answer is B.
Because she surveyed only 25 students (a subset of 600), this is a sample. Use s with denominator n−1.
— σ (sigma) = population parameter, uses N
— s = sample statistic, uses n−1 (Bessel's correction, reduces bias)
— μ (mu) = population mean; x̄ (x-bar) = sample mean
Because she surveyed only 25 students (a subset of 600), this is a sample. Use s with denominator n−1.
— σ (sigma) = population parameter, uses N
— s = sample statistic, uses n−1 (Bessel's correction, reduces bias)
— μ (mu) = population mean; x̄ (x-bar) = sample mean
07
A box plot for exam scores shows:
Minimum = 42, Q1 = 58, Median = 71, Q3 = 82, Maximum = 96
What percentage of students scored between 58 and 82?
Minimum = 42, Q1 = 58, Median = 71, Q3 = 82, Maximum = 96
What percentage of students scored between 58 and 82?
✓ The answer is B — 50%.
The range Q1 to Q3 (the box itself) always contains the middle 50% of data.
Q1 to median = 25%, median to Q3 = 25% → Q1 to Q3 = 50%.
Memory: The BOX in a box plot = 50% of data (from Q1 to Q3).
The range Q1 to Q3 (the box itself) always contains the middle 50% of data.
Q1 to median = 25%, median to Q3 = 25% → Q1 to Q3 = 50%.
Memory: The BOX in a box plot = 50% of data (from Q1 to Q3).
08
The heights of adult men in a city are normally distributed with a mean of 70 inches and a standard deviation of 3 inches.
Using the Empirical Rule, approximately what percentage of men are between 64 and 76 inches tall?
Using the Empirical Rule, approximately what percentage of men are between 64 and 76 inches tall?
✓ The answer is C — 95%.
64 = 70 − 6 = μ − 2σ; 76 = 70 + 6 = μ + 2σ.
The range μ ± 2σ covers 95% of the data by the Empirical Rule (68-95-99.7).
Memory: 1σ → 68%, 2σ → 95%, 3σ → 99.7%
64 = 70 − 6 = μ − 2σ; 76 = 70 + 6 = μ + 2σ.
The range μ ± 2σ covers 95% of the data by the Empirical Rule (68-95-99.7).
Memory: 1σ → 68%, 2σ → 95%, 3σ → 99.7%
09
SAT scores are normally distributed with μ = 1050 and σ = 200.
A student scores 1350.
What is the student's z-score?
A student scores 1350.
What is the student's z-score?
✓ The answer is B — z = 1.5.
z = (x − μ) / σ = (1350 − 1050) / 200 = 300 / 200 = 1.5
Interpretation: The student scored 1.5 standard deviations above the mean.
Memory: z = (x − μ) / σ. Positive z = above mean, negative z = below mean.
z = (x − μ) / σ = (1350 − 1050) / 200 = 300 / 200 = 1.5
Interpretation: The student scored 1.5 standard deviations above the mean.
Memory: z = (x − μ) / σ. Positive z = above mean, negative z = below mean.
10
A dataset has a mean of 45 and a median of 52.
What does this tell you about the distribution's shape?
What does this tell you about the distribution's shape?
✓ The answer is A — Skewed left.
Mean (45) < Median (52) → the mean is pulled LEFT by low outliers → left skew (negative skew).
The tail points to the LEFT.
Memory: Mean chases the tail. Tail left → mean < median. Tail right → mean > median.
Mean (45) < Median (52) → the mean is pulled LEFT by low outliers → left skew (negative skew).
The tail points to the LEFT.
Memory: Mean chases the tail. Tail left → mean < median. Tail right → mean > median.
11
Five values: 2, 4, 4, 4, 6. The mean is 4.
What is the population standard deviation σ?
What is the population standard deviation σ?
Steps: find each (x−μ)², sum them, divide by N, take √
✓ The answer is B.
Deviations from mean (4): (2−4)²=4, (4−4)²=0, (4−4)²=0, (4−4)²=0, (6−4)²=4.
Sum = 4+0+0+0+4 = 8.
Variance σ² = 8/5 = 1.6 (divide by N=5 for population).
σ = √1.6 ≈ 1.265.
Note: if sample SD, divide by n−1=4 → s = √(8/4) = √2 ≈ 1.414. A common mistake!
Deviations from mean (4): (2−4)²=4, (4−4)²=0, (4−4)²=0, (4−4)²=0, (6−4)²=4.
Sum = 4+0+0+0+4 = 8.
Variance σ² = 8/5 = 1.6 (divide by N=5 for population).
σ = √1.6 ≈ 1.265.
Note: if sample SD, divide by n−1=4 → s = √(8/4) = √2 ≈ 1.414. A common mistake!
12
Word Problem
Maya and Jordan take different standardized tests. Maya scores 680 on a test where the mean is 600 and the standard deviation is 80. Jordan scores 27 on a test where the mean is 21 and the standard deviation is 4. Both want to apply to the same university, which compares applicants using standardized scores.
Who performed better relative to their respective test-taking populations?
Who performed better relative to their respective test-taking populations?
✓ The answer is B — Jordan performed better relatively.
Maya's z = (680 − 600) / 80 = 80/80 = 1.0
Jordan's z = (27 − 21) / 4 = 6/4 = 1.5
Jordan is 1.5 standard deviations above his mean; Maya is only 1.0 above hers.
Raw scores on different tests are meaningless for comparison — always convert to z-scores first!
Maya's z = (680 − 600) / 80 = 80/80 = 1.0
Jordan's z = (27 − 21) / 4 = 6/4 = 1.5
Jordan is 1.5 standard deviations above his mean; Maya is only 1.0 above hers.
Raw scores on different tests are meaningless for comparison — always convert to z-scores first!
13
Word Problem
A factory produces bolts whose lengths follow a normal distribution with mean μ = 50 mm and standard deviation σ = 2 mm. Quality control rejects any bolt shorter than 46 mm or longer than 54 mm.
Approximately what percentage of bolts pass quality control?
Approximately what percentage of bolts pass quality control?
✓ The answer is C — 95%.
46 = 50 − 4 = μ − 2σ and 54 = 50 + 4 = μ + 2σ.
By the Empirical Rule, μ ± 2σ contains 95% of data.
So approximately 95% of bolts are within specs and pass QC.
The remaining 5% are rejected (2.5% too short, 2.5% too long).
46 = 50 − 4 = μ − 2σ and 54 = 50 + 4 = μ + 2σ.
By the Empirical Rule, μ ± 2σ contains 95% of data.
So approximately 95% of bolts are within specs and pass QC.
The remaining 5% are rejected (2.5% too short, 2.5% too long).
14
Two classes take the same exam. Their box plots show:
Class A: Min=50, Q1=62, Median=70, Q3=78, Max=90
Class B: Min=40, Q1=55, Median=70, Q3=85, Max=98
Which statement is correct?
Class A: Min=50, Q1=62, Median=70, Q3=78, Max=90
Class B: Min=40, Q1=55, Median=70, Q3=85, Max=98
Which statement is correct?
✓ The answer is D.
Class A IQR = 78 − 62 = 16. Class B IQR = 85 − 55 = 30.
Both have median = 70 (same center), but Class B has a larger IQR (30 > 16), meaning greater spread/variability.
Equal medians ≠ equal distributions. Always check SPREAD (IQR) separately from CENTER (median).
Class A IQR = 78 − 62 = 16. Class B IQR = 85 − 55 = 30.
Both have median = 70 (same center), but Class B has a larger IQR (30 > 16), meaning greater spread/variability.
Equal medians ≠ equal distributions. Always check SPREAD (IQR) separately from CENTER (median).
15
A teacher curves a test by adding 10 points to every student's score.
Which of the following does NOT change after adding 10 points?
Which of the following does NOT change after adding 10 points?
⚠️ This is a classic trap question — many students get this wrong!
✓ The answer is C — Standard deviation does NOT change.
Adding a constant to every value shifts the entire distribution — mean, median, max all increase by 10.
But the spread (how far apart the values are from each other) stays exactly the same!
Standard deviation measures distance between data points → adding a constant doesn't change distances.
Rule: Adding/subtracting a constant → changes mean & median, but NOT standard deviation or IQR.
Adding a constant to every value shifts the entire distribution — mean, median, max all increase by 10.
But the spread (how far apart the values are from each other) stays exactly the same!
Standard deviation measures distance between data points → adding a constant doesn't change distances.
Rule: Adding/subtracting a constant → changes mean & median, but NOT standard deviation or IQR.
16
Word Problem
The battery life of a brand of laptop is normally distributed with a mean of 8 hours and standard deviation of 1.2 hours. A reviewer randomly selects one laptop.
What is the probability that the laptop's battery lasts more than 10.4 hours?
What is the probability that the laptop's battery lasts more than 10.4 hours?
✓ The answer is A — approximately 2.5%.
z = (10.4 − 8) / 1.2 = 2.4 / 1.2 = 2.0
So 10.4 hours is exactly 2σ above the mean.
By Empirical Rule: 95% of data falls within μ ± 2σ → 5% is outside.
Since the normal curve is symmetric: 5% ÷ 2 = 2.5% above μ + 2σ.
Beyond 2σ on one side = 2.5%
z = (10.4 − 8) / 1.2 = 2.4 / 1.2 = 2.0
So 10.4 hours is exactly 2σ above the mean.
By Empirical Rule: 95% of data falls within μ ± 2σ → 5% is outside.
Since the normal curve is symmetric: 5% ÷ 2 = 2.5% above μ + 2σ.
Beyond 2σ on one side = 2.5%
17
A dataset has mean = 50 and standard deviation = 8.
Every value in the dataset is multiplied by 3.
What are the new mean and standard deviation?
Every value in the dataset is multiplied by 3.
What are the new mean and standard deviation?
✓ The answer is C — Mean = 150, SD = 24.
When multiplying by a constant k:
• New mean = k × old mean = 3 × 50 = 150
• New SD = k × old SD = 3 × 8 = 24
(Variance multiplies by k² = 9, but SD multiplies by k = 3)
Rule: Multiplying by k → BOTH mean and SD scale by k. Compare to adding a constant (SD unchanged).
When multiplying by a constant k:
• New mean = k × old mean = 3 × 50 = 150
• New SD = k × old SD = 3 × 8 = 24
(Variance multiplies by k² = 9, but SD multiplies by k = 3)
Rule: Multiplying by k → BOTH mean and SD scale by k. Compare to adding a constant (SD unchanged).
18
Word Problem
IQ scores are normally distributed with μ = 100 and σ = 15. A gifted program accepts students who score in the top 2.5% of the population. Using the Empirical Rule, what is the minimum IQ score needed to qualify for the gifted program?
✓ The answer is B — 130.
Top 2.5% means above μ + 2σ (from Empirical Rule: 95% within 2σ → 5% outside → 2.5% in each tail).
μ + 2σ = 100 + 2(15) = 100 + 30 = 130.
So a student needs at least IQ = 130 to be in the top 2.5%.
Reverse z-score: x = μ + z·σ. Here z = 2 (top 2.5%).
Top 2.5% means above μ + 2σ (from Empirical Rule: 95% within 2σ → 5% outside → 2.5% in each tail).
μ + 2σ = 100 + 2(15) = 100 + 30 = 130.
So a student needs at least IQ = 130 to be in the top 2.5%.
Reverse z-score: x = μ + z·σ. Here z = 2 (top 2.5%).
19
Word Problem
Two coffee shops record their daily sales (in dollars) over one week:
Shop A: 420, 430, 425, 415, 435, 440, 425
Shop B: 200, 350, 500, 600, 150, 700, 450
Both shops have approximately the same mean (~427 vs ~421). A business analyst wants to determine which shop has more consistent sales. Which measure should she use, and which shop is more consistent?
Shop A: 420, 430, 425, 415, 435, 440, 425
Shop B: 200, 350, 500, 600, 150, 700, 450
Both shops have approximately the same mean (~427 vs ~421). A business analyst wants to determine which shop has more consistent sales. Which measure should she use, and which shop is more consistent?
✓ The answer is C.
"Consistency" = low variability → use standard deviation (or IQR).
Shop A's values cluster tightly around the mean (SD ≈ 8). Shop B swings wildly from 150 to 700 (SD ≈ 190).
Same mean can hide very different spreads — always report both center AND spread.
Note: Range works too, but SD is preferred as it uses all data points, not just min & max.
"Consistency" = low variability → use standard deviation (or IQR).
Shop A's values cluster tightly around the mean (SD ≈ 8). Shop B swings wildly from 150 to 700 (SD ≈ 190).
Same mean can hide very different spreads — always report both center AND spread.
Note: Range works too, but SD is preferred as it uses all data points, not just min & max.
20
Final Challenge — Word Problem
A high school athletic trainer records sprint times (in seconds) for 9 athletes on the track team:
11.2, 10.8, 11.5, 11.0, 10.9, 11.3, 11.1, 10.7, 15.2
The trainer wants to report a measure of center that is not distorted by the athlete who ran 15.2 seconds (clearly injured during the trial). She also wants to report a measure of spread that is resistant to the same outlier.
Which combination should she use?
11.2, 10.8, 11.5, 11.0, 10.9, 11.3, 11.1, 10.7, 15.2
The trainer wants to report a measure of center that is not distorted by the athlete who ran 15.2 seconds (clearly injured during the trial). She also wants to report a measure of spread that is resistant to the same outlier.
Which combination should she use?
✓ The answer is B — Median and IQR.
The 15.2 second time is a clear outlier that pulls the mean upward and inflates standard deviation.
Without outlier: mean ≈ 11.06 s. With outlier: mean ≈ 11.63 s — a big shift!
The median (sort: 10.7, 10.8, 10.9, 11.0, 11.1, 11.2, 11.3, 11.5, 15.2 → median = 11.1) is barely affected.
The IQR also ignores extreme values by focusing on the middle 50%.
Golden Rule: Outliers present → use MEDIAN + IQR. Symmetric/no outliers → use MEAN + SD.
The 15.2 second time is a clear outlier that pulls the mean upward and inflates standard deviation.
Without outlier: mean ≈ 11.06 s. With outlier: mean ≈ 11.63 s — a big shift!
The median (sort: 10.7, 10.8, 10.9, 11.0, 11.1, 11.2, 11.3, 11.5, 15.2 → median = 11.1) is barely affected.
The IQR also ignores extreme values by focusing on the middle 50%.
Golden Rule: Outliers present → use MEDIAN + IQR. Symmetric/no outliers → use MEAN + SD.