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Statistics & Data Science MCQs multiple-choice questions with answers & explanations

All 20 Statistics & Data Science quiz questions on one page. Pick an answer in your head, then open Show answer to check it and read why. Want a score and a timer? Take them as a quiz instead.

Written by Aditya MishraTechnical Lead, BNP Paribas

Official reference: NIST Engineering Statistics Handbook

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20 questions

1.
Nine customers spend 20 and one spends 2,000. What does this print?
easy
```
from statistics import mean, median
spend = [20] * 9 + [2000]
print(mean(spend), median(spend))
```
1. A218 20.0
2. B218 218
3. C20 20.0
4. D218.0 1010.0
Show answer
Answer: A (218 20.0)
The mean is 2,180 / 10 = 218 and stays an int because every input is an int. The median of an even-length list averages the two middle values, 20 and 20, so it returns the float 20.0. One outlier moves the mean tenfold and leaves the median untouched.
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2.
What does this print?
easy
```
from statistics import pvariance, variance
data = [2, 4, 4, 4, 5, 5, 7, 9]
print(pvariance(data), round(variance(data), 3))
```
1. A4 4
2. B4 4.571
3. C4.571 4
4. D2 2.138
Show answer
Answer: B (4 4.571)
The squared deviations from the mean of 5 sum to 32. pvariance divides by n = 8, giving 4; variance is the sample variance and divides by n - 1 = 7, giving 4.571. The n - 1 version corrects for measuring deviations from the sample mean.
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3.
Scores are normal with mean 100 and standard deviation 15. What does this print?
easy
```
from statistics import NormalDist
iq = NormalDist(mu=100, sigma=15)
print(round(iq.cdf(130), 3), round(iq.inv_cdf(0.975), 1))
```
1. A0.95 130.0
2. B0.977 129.4
3. C0.997 145.0
4. D0.841 115.0
Show answer
Answer: B (0.977 129.4)
130 is two standard deviations above the mean, and about 97.7% of a normal distribution lies below z = 2. The 97.5th percentile is 1.96 standard deviations up, 100 + 1.96 * 15 = 129.4, not exactly two.
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4.
A condition has 1% prevalence; the test has 95% sensitivity and 95% specificity. What does this print?
mid
```
prior, sens, spec = 0.01, 0.95, 0.95
p_pos = sens * prior + (1 - spec) * (1 - prior)
print(round(sens * prior / p_pos, 3))
```
1. A0.95
2. B0.5
3. C0.161
4. D0.01
Show answer
Answer: C (0.161)
Of 10,000 people, 95 sick people test positive and 495 healthy people also test positive, so only 95 / 590 = 0.161 of positives are real. Answering 0.95 confuses the sensitivity P(+ | sick) with the posterior P(sick | +) and ignores the base rate.
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5.
An experiment tracks 20 independent metrics, none of which is truly affected. At alpha 0.05, what does this print?
mid
```
print(round(1 - 0.95 ** 20, 2))
```
1. A0.05
2. B0.36
3. C0.64
4. D1.0
Show answer
Answer: C (0.64)
The chance that every one of 20 null tests stays non-significant is 0.95^20, about 0.36, so the chance of at least one false positive is about 0.64. That is why experiments name one primary metric in advance or correct for multiple comparisons.
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6.
A service averages 3 errors per hour (Poisson). What does this print?
mid
```
import math
lam = 3
print(round(math.exp(-lam), 3),
      round(1 - math.exp(-lam) * sum(lam**k / math.factorial(k) for k in range(3)), 3))
```
1. A0.05 0.577
2. B0.0 0.5
3. C0.05 0.423
4. D0.333 0.667
Show answer
Answer: A (0.05 0.577)
For a Poisson with rate 3, P(0) = e^-3, about 0.05. The second value is P(X >= 3) = 1 - P(0) - P(1) - P(2) = 1 - e^-3 (1 + 3 + 4.5), about 0.577. 0.423 is P(X <= 2), the complement.
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7.
What is the probability of at least one six in four rolls of a fair die?
easy
```
print(round(1 - (5 / 6) ** 4, 3))
```
1. A0.667
2. B0.518
3. C0.482
4. D0.167
Show answer
Answer: B (0.518)
Use the complement: the chance of no six in four independent rolls is (5/6)^4, about 0.482, so at least one six is 0.518. Adding 1/6 four times (0.667) double-counts outcomes with several sixes.
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8.
What does this birthday-problem calculation print for 23 people?
mid
```
p_unique = 1.0
for i in range(23):
    p_unique *= (365 - i) / 365
print(round(1 - p_unique, 3))
```
1. A0.063
2. B0.23
3. C0.507
4. D0.999
Show answer
Answer: C (0.507)
The loop multiplies the chances that each new person avoids every earlier birthday, and 23 people already give 253 pairs, so a shared birthday is slightly more likely than not. 0.063 is the chance that someone shares a birthday with one specific person, a different question.
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9.
In Monty Hall, the host always opens a goat door you did not pick. What win rate for switching does this print?
mid
```
import random
rng = random.Random(42)
wins = 0
for _ in range(100_000):
    car, pick = rng.randrange(3), rng.randrange(3)
    wins += pick != car   # switching wins exactly when the first pick was wrong
print(round(wins / 100_000, 2))
```
1. A0.33
2. B0.5
3. C0.67
4. D1.0
Show answer
Answer: C (0.67)
Your first pick is wrong two times in three. Because the host knowingly removes the other goat, switching turns every wrong first pick into a win, so it wins about 2/3 of the time. The 0.5 intuition treats the two remaining doors as symmetric, but the host's choice depended on where the car was.
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10.
Draws come from a skewed exponential distribution with mean 1 and standard deviation 1. What does this print?
mid
```
import random
from statistics import mean, stdev
rng = random.Random(0)
means = [mean(rng.expovariate(1.0) for _ in range(30)) for _ in range(2000)]
print(round(mean(means), 2), round(stdev(means), 2))
```
1. A1.0 1.0
2. B1.0 0.18
3. C0.69 0.18
4. D1.0 0.03
Show answer
Answer: B (1.0 0.18)
The sample means centre on the population mean, 1.0, and their spread is the standard error sigma / sqrt(n) = 1 / sqrt(30), about 0.18. The spread of the raw data (1.0) is not the spread of means, and 0.69 is the median of the exponential, not the mean.
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11.
What does this print?
mid
```
from statistics import correlation
x = [-3, -2, -1, 0, 1, 2, 3]
y = [v * v for v in x]
print(correlation(x, y))
```
1. A1.0
2. B0.0
3. C-1.0
4. D0.5
Show answer
Answer: B (0.0)
Pearson correlation measures linear association only. y is completely determined by x, but the relationship is a symmetric parabola, so the positive and negative halves cancel and the correlation is exactly 0. Zero correlation does not mean independence.
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12.
What does this print?
easy
```
from statistics import linear_regression
slope, intercept = linear_regression([1, 2, 3, 4, 5], [2, 4, 5, 4, 5])
print(round(slope, 2), round(intercept, 2))
```
1. A1.0 1.0
2. B0.6 2.2
3. C0.75 1.5
4. D0.6 4.0
Show answer
Answer: B (0.6 2.2)
The least-squares slope is cov(x, y) / var(x) = 1.5 / 2.5 = 0.6, and the line passes through the means (3, 4), so the intercept is 4 - 0.6 * 3 = 2.2. statistics.linear_regression returns a named tuple you can unpack.
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13.
Two onboarding flows got different device mixes. What does the last value on each line show?
mid
```
data = {  # (conversions, users)
    "old": {"mobile": (20, 400), "desktop": (90, 600)},
    "new": {"mobile": (60, 1000), "desktop": (19, 100)},
}
for flow, seg in data.items():
    rates = {k: c / n for k, (c, n) in seg.items()}
    total = sum(c for c, _ in seg.values()) / sum(n for _, n in seg.values())
    print(flow, {k: round(v, 3) for k, v in rates.items()}, round(total, 3))
# old {'mobile': 0.05, 'desktop': 0.15} 0.11
# new {'mobile': 0.06, 'desktop': 0.19} 0.072
```
1. AThe new flow wins overall, matching each segment
2. BThe new flow wins in both segments but loses overall
3. CThe new flow loses in both segments and overall
4. DThe overall rate is the average of the two segment rates
Show answer
Answer: B (The new flow wins in both segments but loses overall)
The new flow converts better on mobile (6% vs 5%) and desktop (19% vs 15%), but 91% of its users are on mobile, which converts poorly, so its pooled rate is lower. That reversal is Simpson's paradox; the overall rate is a mix-weighted average, not the simple average of segments.
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14.
Control converts 100 of 1,000 and treatment 130 of 1,000. What does this print?
hard
```
import math
c1, n1, c2, n2 = 100, 1000, 130, 1000
p = (c1 + c2) / (n1 + n2)
z = (c2 / n2 - c1 / n1) / math.sqrt(p * (1 - p) * (1 / n1 + 1 / n2))
obs = [c1, n1 - c1, c2, n2 - c2]
exp = [n1 * p, n1 * (1 - p), n2 * p, n2 * (1 - p)]
chi2 = sum((o - e) ** 2 / e for o, e in zip(obs, exp))
print(round(z, 3), round(z * z, 3), round(chi2, 3))
```
1. A2.103 4.422 4.422
2. B2.103 4.422 2.103
3. C1.96 3.841 3.841
4. D2.103 4.422 8.844
Show answer
Answer: A (2.103 4.422 4.422)
The pooled two-proportion z statistic is 2.103, and the Pearson chi-square statistic on the same 2x2 table (without continuity correction) is exactly z squared, 4.422. Both give the same two-sided p-value, about 0.035.
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15.
Waiting times are exponential with rate 0.5 per minute. What does this print?
mid
```
import math
rate = 0.5
p_gt = lambda t: math.exp(-rate * t)
print(round(p_gt(3) / p_gt(1), 3), round(p_gt(2), 3))
```
1. A0.223 0.368
2. B0.368 0.368
3. C0.607 0.368
4. D0.368 0.135
Show answer
Answer: B (0.368 0.368)
The first value is P(T > 3 | T > 1), and it equals P(T > 2) = e^-1, about 0.368. That is the memoryless property: having already waited one minute does not change the distribution of the remaining wait.
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16.
What does this print?
hard
```
from statistics import quantiles
print(quantiles(range(1, 11), n=4))
```
1. A[3, 5.5, 8]
2. B[2.75, 5.5, 8.25]
3. C[3.25, 5.5, 7.75]
4. D[2.5, 5, 7.5]
Show answer
Answer: B ([2.75, 5.5, 8.25])
statistics.quantiles defaults to method="exclusive", which places cut points at positions (n + 1)p and gives 2.75 and 8.25 for the quartiles. method="inclusive" (the convention NumPy uses by default) gives 3.25 and 7.75. Different tools disagree on small samples, so state the method.
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17.
User 1 has orders of 50, 50 and 20. For the 20 order, what are rn, rnk and drnk?
mid
```
SELECT user_id, amount,
  ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY amount DESC) AS rn,
  RANK()       OVER (PARTITION BY user_id ORDER BY amount DESC) AS rnk,
  DENSE_RANK() OVER (PARTITION BY user_id ORDER BY amount DESC) AS drnk
FROM orders;
```
1. A3, 3, 3
2. B3, 2, 2
3. C3, 3, 2
4. D2, 3, 2
Show answer
Answer: C (3, 3, 2)
The two 50s tie. ROW_NUMBER still numbers rows 1, 2, 3; RANK gives both ties 1 and skips to 3; DENSE_RANK gives both ties 1 and continues with 2. Pick ROW_NUMBER when you need exactly one row per user, such as the first order.
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18.
An A/A test (no real difference) is checked daily for 20 days and stopped the first time p is below 0.05. Roughly what is the false-positive rate?
mid
1. A5%, because alpha is 0.05
2. BAbout 2.5%, because only one direction can win
3. CAround 20 to 25%
4. D100%, given enough days
Show answer
Answer: C (Around 20 to 25%)
Every look gives noise another chance to cross the threshold, and stopping at the first crossing locks in the false positive. A simulation with 20 looks gives about 24%, not 5%. It approaches 100% only with unlimited looks; sequential methods fix the problem by adjusting the boundaries.
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19.
A 95% confidence interval for a conversion rate is plus or minus 2 points with 1,000 users. About how wide is it with 4,000 users?
easy
1. APlus or minus 0.5 points
2. BPlus or minus 1 point
3. CPlus or minus 2 points
4. DPlus or minus 4 points
Show answer
Answer: B (Plus or minus 1 point)
The margin of error is proportional to the standard error, which shrinks with 1 / sqrt(n). Four times the users halves the margin, so it is about 1 point. Expecting it to shrink by four confuses n with sqrt(n).
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20.
Which change increases the power of an A/B test, everything else fixed?
mid
1. ALowering alpha from 0.05 to 0.01
2. BChoosing a smaller minimum detectable effect
3. CUsing CUPED to reduce the metric's variance
4. DSplitting traffic 90/10 instead of 50/50
Show answer
Answer: C (Using CUPED to reduce the metric's variance)
Power rises with lower variance, larger effects, more data and looser alpha. CUPED removes pre-existing variation, so the same effect stands out more clearly. A stricter alpha, a smaller target effect and an unbalanced split all reduce power for a given total sample.
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