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Continuous Probability Distributions MCQs 2026

50 questions with detailed answers · 34 from past papers · 5 quiz batches available

📚 Statistics Mcqs 📄 34 Past-Paper Qs ✓ Free · No Login Needed
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Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1 Questions 1–10 of 50
  1. Q1 medium

    For continuous uniform on [a,b], P(X < c) for a < c < b equals

    1. A c/b
    2. B (c − a)/(b − a)
    3. C 1/(b−a)
    4. D 0.5 always
    💡 Explanation:

    CDF of uniform is linear: (c−a)/(b−a).

  2. Q2 hard

    The 90th percentile of Z is approximately

    1. A 1.96
    2. B 0.90
    3. C 1.28
    4. D 2.58
    💡 Explanation:

    P(Z≤1.28)≈0.90.

  3. Q3 Past Paper · PPSC/FPSC/CSS medium

    For manufacturing: bolt diameter ~ N(10, 0.01). σ=0.1 mm. P(diameter < 9.8) uses Z=

    1. A −0.2
    2. B −2
    3. C 2
    4. D 0
    💡 Explanation:

    Z=(9.8−10)/0.1=−2.

  4. Q4 Past Paper · PPSC/FPSC/CSS easy

    The standard normal distribution Z has

    1. A mean 0 and variance 1
    2. B mean 1 and variance 0
    3. C mean μ and variance σ
    4. D uniform on 0 to 1
    💡 Explanation:

    Z ~ N(0,1) is the reference normal.

  5. Q5 Past Paper · PPSC/FPSC/CSS easy

    If X ~ N(μ, σ²), the standardized variable Z equals

    1. A X / μ
    2. B (X − σ) / μ
    3. C μ / σ
    4. D (X − μ) / σ
    💡 Explanation:

    Z-score measures deviation in standard deviation units.

  6. Q6 Past Paper · PPSC/FPSC/CSS easy

    For X ~ N(100, 25), σ equals

    1. A 25
    2. B 5
    3. C 100
    4. D 12.5
    💡 Explanation:

    σ = √25 = 5.

  7. Q7 Past Paper · PPSC/FPSC/CSS easy

    For X ~ N(100, 25), P(Z=0) corresponds to X equal to

    1. A 100
    2. B 25
    3. C 5
    4. D 0
    💡 Explanation:

    Z=0 when X=μ=100.

  8. Q8 Past Paper · PPSC/FPSC/CSS easy

    The normal curve is

    1. A bell-shaped and symmetric about μ
    2. B skewed right always
    3. C uniform
    4. D discrete
    💡 Explanation:

    Normal PDF is symmetric unimodal bell curve.

  9. Q9 Past Paper · PPSC/FPSC/CSS easy

    Total area under the normal PDF equals

    1. A 1
    2. B 0.5
    3. C 0
    4. D μ
    💡 Explanation:

    Any valid PDF integrates to 1.

  10. Q10 Past Paper · PPSC/FPSC/CSS easy

    For standard normal Z, approximately 68% of probability lies within

    1. A Z = ±2
    2. B Z = ±3
    3. C Z = ±0.5
    4. D Z = ±1
    💡 Explanation:

    Empirical rule: ~68% within 1σ.

  11. Q11 Past Paper · PPSC/FPSC/CSS easy

    Approximately 95% of a normal distribution lies within

    1. A μ ± 2σ
    2. B μ ± 1σ
    3. C μ ± 3σ
    4. D μ only
    💡 Explanation:

    Empirical rule: ~95% within 2σ.

  12. Q12 Past Paper · PPSC/FPSC/CSS easy

    Approximately 99.7% lies within

    1. A μ ± 3σ
    2. B μ ± 2σ
    3. C μ ± 1σ
    4. D μ ± 4σ
    💡 Explanation:

    Empirical rule: nearly all within 3σ.

  13. Q13 Past Paper · PPSC/FPSC/CSS easy

    P(Z ≤ 0) for standard normal equals

    1. A 0
    2. B 0.5
    3. C 1
    4. D 0.68
    💡 Explanation:

    Symmetry: half below mean 0.

  14. Q14 Past Paper · PPSC/FPSC/CSS medium

    If X ~ N(50, 100), P(X < 60) uses Z equal to

    1. A 1
    2. B 0.5
    3. C 2
    4. D 10
    💡 Explanation:

    Z = (60−50)/10 = 1.

  15. Q15 Past Paper · PPSC/FPSC/CSS medium

    If X ~ N(200, 64), P(X > 216) uses Z equal to

    1. A 1
    2. B 4
    3. C 2
    4. D 16
    💡 Explanation:

    Z = (216−200)/8 = 2.

  16. Q16 Past Paper · PPSC/FPSC/CSS medium

    P(Z > 1.96) is approximately

    1. A 0.05
    2. B 0.5
    3. C 0.025
    4. D 0.975
    💡 Explanation:

    Upper 2.5% tail at ±1.96 for 95% central.

  17. Q17 Past Paper · PPSC/FPSC/CSS medium

    A 95% confidence interval for μ uses critical value approximately

    1. A ±1.00 always
    2. B ±1.96 for large samples (known σ)
    3. C ±3.00 always
    4. D 0
    💡 Explanation:

    Standard 95% two-tailed z critical is 1.96.

  18. Q18 easy

    The normal PDF f(x) for X ~ N(μ,σ²) is highest at

    1. A x = σ
    2. B x = 0 always
    3. C x = μ
    4. D x = μ + σ
    💡 Explanation:

    Mode and mean coincide at μ for normal.

  19. Q19 easy

    For normal X, median and mean are

    1. A median always above mean
    2. B equal (both μ)
    3. C mean always zero
    4. D unequal always
    💡 Explanation:

    Symmetric distribution: mean = median = mode.

  20. Q20 medium

    The uniform distribution on [a,b] has PDF

    1. A b−a everywhere
    2. B a constant > 1 always
    3. C 1/(b−a) on [a,b] and 0 elsewhere
    4. D zero on [a,b]
    💡 Explanation:

    Uniform density is constant over the interval.

  21. Q21 Past Paper · PPSC/FPSC/CSS easy

    Uniform on [0,10]: E(X) equals

    1. A 10
    2. B 2.5
    3. C 0
    4. D 5
    💡 Explanation:

    E(X)=(a+b)/2=5.

  22. Q22 medium

    Uniform on [0,10]: Var(X) equals

    1. A 25
    2. B 100/12 ≈ 8.333
    3. C 10
    4. D 5
    💡 Explanation:

    Var=(b−a)²/12=100/12.

  23. Q23 Past Paper · PPSC/FPSC/CSS medium

    The exponential distribution is often used for

    1. A waiting time until an event in a Poisson process
    2. B counting discrete successes in n trials
    3. C sampling without replacement
    4. D hypergeometric counts
    💡 Explanation:

    Exponential models continuous waiting times with memoryless property.

  24. Q24 medium

    Exponential(λ) has mean

    1. A λ
    2. B λ²
    3. C 0
    4. D 1/λ
    💡 Explanation:

    Mean waiting time is 1/λ.

  25. Q25 hard

    The memoryless property holds for

    1. A normal distribution
    2. B uniform only
    3. C binomial
    4. D exponential (and geometric discrete counterpart)
    💡 Explanation:

    P(X>s+t|X>s)=P(X>t) for exponential.

  26. Q26 Past Paper · PPSC/FPSC/CSS medium

    Normal approximation to binomial applies when

    1. A n is very small
    2. B p is exactly 0
    3. C np and n(1−p) are both sufficiently large (often ≥ 5 or 10)
    4. D np is near 0 only
    💡 Explanation:

    CLT/de Moivre–Laplace: large n with moderate p.

  27. Q27 Past Paper · PPSC/FPSC/CSS medium

    For X ~ Bin(100, 0.5), normal approximation uses μ and σ² equal to

    1. A 100 and 0.5
    2. B 50 and 50
    3. C 25 and 50
    4. D 50 and 25
    💡 Explanation:

    μ=np=50; σ²=np(1−p)=25.

  28. Q28 Past Paper · PPSC/FPSC/CSS medium

    Continuity correction when approximating P(X=k) for discrete X uses

    1. A P(Y=k) exactly
    2. B P(Y<k) only without adjustment
    3. C P(k−0.5 < Y < k+0.5) for normal Y
    4. D no correction ever
    💡 Explanation:

    Half-unit adjustment bridges discrete and continuous.

  29. Q29 Past Paper · PPSC/FPSC/CSS hard

    Bin(100,0.5): approximate P(X ≤ 45) with continuity correction uses

    1. A P(Y < 45.5) for Y ~ N(50,25)
    2. B P(Y < 45)
    3. C P(Y < 50)
    4. D P(Y < 40.5)
    💡 Explanation:

    P(X≤45) ≈ P(Y≤45.5) with correction.

  30. Q30 hard

    For Bin(400, 0.25), normal approx: μ=100, σ=√75≈8.66. Z for X=110 uses

    1. A (110−100)/8.66 only
    2. B 100/8.66
    3. C (110−100)/75
    4. D (110.5−100)/8.66 with continuity correction
    💡 Explanation:

    Use 110.5 for P(X≤110); σ=√(np(1−p)).

  31. Q31 Past Paper · PPSC/FPSC/CSS medium

    Poisson(λ) can be approximated by N(λ, λ) when

    1. A λ = 0
    2. B λ is large (e.g., λ ≥ 10)
    3. C λ = 1 always
    4. D λ < 1 only
    💡 Explanation:

    Large λ: Poisson approaches normal.

  32. Q32 hard

    For Poisson(100), approximate P(X ≥ 115) uses normal with μ=100, σ=10 and

    1. A no correction
    2. B μ=50
    3. C continuity correction at 114.5
    4. D σ=100
    💡 Explanation:

    σ=√100=10; P(X≥115)≈P(Y>114.5).

  33. Q33 Past Paper · PPSC/FPSC/CSS easy

    The standard normal table typically gives

    1. A P(Z = z)
    2. B P(Z ≥ z) only without conversion
    3. C P(X = μ)
    4. D P(Z ≤ z)
    💡 Explanation:

    Tables tabulate left-tail cumulative probabilities.

  34. Q34 Past Paper · PPSC/FPSC/CSS easy

    To find P(Z > a), compute

    1. A P(Z ≤ a)
    2. B 1 − P(Z ≤ a)
    3. C P(Z ≤ −a)
    4. D P(Z = a)
    💡 Explanation:

    Upper tail = 1 minus CDF.

  35. Q35 Past Paper · PPSC/FPSC/CSS medium

    P(−1.5 < Z < 1.5) equals

    1. A P(Z<1.5) + P(Z<−1.5)
    2. B 1 − P(Z<1.5)
    3. C P(Z<1.5) − P(Z<−1.5)
    4. D P(Z<3)
    💡 Explanation:

    Interval probability = difference of CDF values.

  36. Q36 Past Paper · PPSC/FPSC/CSS easy

    By symmetry P(Z < −a) equals

    1. A P(Z > a)
    2. B P(Z < a)
    3. C 0
    4. D 1 − P(Z < 0)
    💡 Explanation:

    Standard normal is symmetric about 0.

  37. Q37 hard

    If X ~ N(μ, σ²), then aX + b (a>0) is distributed as

    1. A N(μ, σ²)
    2. B N(aμ + b, a²σ²)
    3. C N(μ + b, σ²)
    4. D uniform
    💡 Explanation:

    Linear transform of normal is normal.

  38. Q38 hard

    Sum of independent normal variables is

    1. A always standard normal
    2. B binomial
    3. C Poisson
    4. D normal with mean and variance added
    💡 Explanation:

    Normal family closed under independent sums.

  39. Q39 Past Paper · PPSC/FPSC/CSS medium

    For X ~ N(10, 4), P(8 < X < 12) uses Z bounds

    1. A 0 and 1
    2. B −2 and +2
    3. C 8 and 12
    4. D −1 and +1
    💡 Explanation:

    Z=(8−10)/2=−1 and (12−10)/2=+1.

  40. Q40 Past Paper · PPSC/FPSC/CSS medium

    P(|Z| > 2) is approximately

    1. A 0.0228
    2. B 0.9544
    3. C 0.0456
    4. D 0.5
    💡 Explanation:

    Two tails beyond ±2: ≈0.0456 total.

  41. Q41 Past Paper · PPSC/FPSC/CSS medium

    A test score X ~ N(70, 100). A score of 85 has Z-score

    1. A 15
    2. B 0.15
    3. C 1.5
    4. D 2.5
    💡 Explanation:

    Z=(85−70)/10=1.5.

  42. Q42 Past Paper · PPSC/FPSC/CSS medium

    Heights ~ N(170, 25). Proportion with height > 180 cm uses Z equal to

    1. A 10
    2. B 1
    3. C 2
    4. D 0.4
    💡 Explanation:

    Z=(180−170)/5=2.

  43. Q43 Past Paper · PPSC/FPSC/CSS medium

    Central Limit Theorem states that

    1. A any small sample is exactly normal
    2. B the sample mean is approximately normal for large n (under conditions)
    3. C population must be normal always for any n
    4. D variance equals zero
    💡 Explanation:

    CLT: X̄ₙ ≈ N(μ, σ²/n) for large n.

  44. Q44 Past Paper · PPSC/FPSC/CSS medium

    For non-normal population with large n, distribution of X̄ is approximately

    1. A Bin(n,p)
    2. B Poisson(λ)
    3. C uniform only
    4. D N(μ, σ²/n)
    💡 Explanation:

    CLT applies regardless of population shape if n large.

  45. Q45 Past Paper · PPSC/FPSC/CSS medium

    Standard error of the mean is

    1. A σ / √n
    2. B σ × √n
    3. C σ²
    4. D σ / n
    💡 Explanation:

    SE = σ/√n shrinks with sample size.

  46. Q46 Past Paper · PPSC/FPSC/CSS easy

    If σ=20 and n=100, SE equals

    1. A 20
    2. B 0.2
    3. C 200
    4. D 2
    💡 Explanation:

    SE = 20/√100 = 2.

  47. Q47 hard

    Normal probability plot is used to

    1. A compute binomial coefficients
    2. B assess whether data plausibly come from a normal distribution
    3. C find Poisson λ exactly
    4. D prove independence
    💡 Explanation:

    QQ/plot checks normality assumption.

  48. Q48 hard

    Using normal approx for Bin(20, 0.4), μ=8, σ≈2.19. P(X≥10) with correction approximates P(Y>9.5)

    1. A using Z=(9.5−8)/2.19
    2. B Z=(10−8)/2.19 without correction
    3. C Z=8/2.19
    4. D Z=0
    💡 Explanation:

    P(X≥10)=P(X>9.5) with correction → P(Y>9.5).

  49. Q49 medium

    Skewed data with small n should

    1. A not rely blindly on normal approximation
    2. B always use Z=1.96
    3. C ignore continuity correction
    4. D assume σ=0
    💡 Explanation:

    Small n or strong skew violates normal approx assumptions.

  50. Q50 hard

    The PDF of N(μ,σ²) has inflection points at

    1. A μ only
    2. B μ ± σ
    3. C μ ± 2σ
    4. D 0 only
    💡 Explanation:

    Second derivative zero at one standard deviation from mean.

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