Continuous Probability Distributions MCQs 2026
50 questions with detailed answers · 34 from past papers · 5 quiz batches available
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- Q1Past Paper · PPSC/FPSC/CSSmedium
Normal approximation to binomial applies when
💡 Explanation:CLT/de Moivre–Laplace: large n with moderate p.
- Q2Past Paper · PPSC/FPSC/CSSeasy
The standard normal distribution Z has
💡 Explanation:Z ~ N(0,1) is the reference normal.
- Q3Past Paper · PPSC/FPSC/CSSeasy
If X ~ N(μ, σ²), the standardized variable Z equals
💡 Explanation:Z-score measures deviation in standard deviation units.
- Q4Past Paper · PPSC/FPSC/CSSeasy
For X ~ N(100, 25), σ equals
💡 Explanation:σ = √25 = 5.
- Q5Past Paper · PPSC/FPSC/CSSeasy
For X ~ N(100, 25), P(Z=0) corresponds to X equal to
💡 Explanation:Z=0 when X=μ=100.
- Q6Past Paper · PPSC/FPSC/CSSeasy
The normal curve is
💡 Explanation:Normal PDF is symmetric unimodal bell curve.
- Q7Past Paper · PPSC/FPSC/CSSeasy
Total area under the normal PDF equals
💡 Explanation:Any valid PDF integrates to 1.
- Q8Past Paper · PPSC/FPSC/CSSeasy
For standard normal Z, approximately 68% of probability lies within
💡 Explanation:Empirical rule: ~68% within 1σ.
- Q9Past Paper · PPSC/FPSC/CSSeasy
Approximately 95% of a normal distribution lies within
💡 Explanation:Empirical rule: ~95% within 2σ.
- Q10Past Paper · PPSC/FPSC/CSSeasy
Approximately 99.7% lies within
💡 Explanation:Empirical rule: nearly all within 3σ.
- Q11Past Paper · PPSC/FPSC/CSSeasy
P(Z ≤ 0) for standard normal equals
💡 Explanation:Symmetry: half below mean 0.
- Q12Past Paper · PPSC/FPSC/CSSmedium
If X ~ N(50, 100), P(X < 60) uses Z equal to
💡 Explanation:Z = (60−50)/10 = 1.
- Q13Past Paper · PPSC/FPSC/CSSmedium
If X ~ N(200, 64), P(X > 216) uses Z equal to
💡 Explanation:Z = (216−200)/8 = 2.
- Q14Past Paper · PPSC/FPSC/CSSmedium
P(Z > 1.96) is approximately
💡 Explanation:Upper 2.5% tail at ±1.96 for 95% central.
- Q15Past Paper · PPSC/FPSC/CSSmedium
A 95% confidence interval for μ uses critical value approximately
💡 Explanation:Standard 95% two-tailed z critical is 1.96.
- Q16easy
The normal PDF f(x) for X ~ N(μ,σ²) is highest at
💡 Explanation:Mode and mean coincide at μ for normal.
- Q17easy
For normal X, median and mean are
💡 Explanation:Symmetric distribution: mean = median = mode.
- Q18medium
The uniform distribution on [a,b] has PDF
💡 Explanation:Uniform density is constant over the interval.
- Q19Past Paper · PPSC/FPSC/CSSeasy
Uniform on [0,10]: E(X) equals
💡 Explanation:E(X)=(a+b)/2=5.
- Q20medium
Uniform on [0,10]: Var(X) equals
💡 Explanation:Var=(b−a)²/12=100/12.
- Q21Past Paper · PPSC/FPSC/CSSmedium
The exponential distribution is often used for
💡 Explanation:Exponential models continuous waiting times with memoryless property.
- Q22medium
Exponential(λ) has mean
💡 Explanation:Mean waiting time is 1/λ.
- Q23hard
The memoryless property holds for
💡 Explanation:P(X>s+t|X>s)=P(X>t) for exponential.
- Q24Past Paper · PPSC/FPSC/CSSmedium
For X ~ Bin(100, 0.5), normal approximation uses μ and σ² equal to
💡 Explanation:μ=np=50; σ²=np(1−p)=25.
- Q25Past Paper · PPSC/FPSC/CSSmedium
Continuity correction when approximating P(X=k) for discrete X uses
💡 Explanation:Half-unit adjustment bridges discrete and continuous.
- Q26Past Paper · PPSC/FPSC/CSShard
Bin(100,0.5): approximate P(X ≤ 45) with continuity correction uses
💡 Explanation:P(X≤45) ≈ P(Y≤45.5) with correction.
- Q27hard
For Bin(400, 0.25), normal approx: μ=100, σ=√75≈8.66. Z for X=110 uses
💡 Explanation:Use 110.5 for P(X≤110); σ=√(np(1−p)).
- Q28Past Paper · PPSC/FPSC/CSSmedium
Poisson(λ) can be approximated by N(λ, λ) when
💡 Explanation:Large λ: Poisson approaches normal.
- Q29hard
For Poisson(100), approximate P(X ≥ 115) uses normal with μ=100, σ=10 and
💡 Explanation:σ=√100=10; P(X≥115)≈P(Y>114.5).
- Q30Past Paper · PPSC/FPSC/CSSeasy
The standard normal table typically gives
💡 Explanation:Tables tabulate left-tail cumulative probabilities.
- Q31Past Paper · PPSC/FPSC/CSSeasy
To find P(Z > a), compute
💡 Explanation:Upper tail = 1 minus CDF.
- Q32Past Paper · PPSC/FPSC/CSSmedium
P(−1.5 < Z < 1.5) equals
💡 Explanation:Interval probability = difference of CDF values.
- Q33Past Paper · PPSC/FPSC/CSSeasy
By symmetry P(Z < −a) equals
💡 Explanation:Standard normal is symmetric about 0.
- Q34hard
If X ~ N(μ, σ²), then aX + b (a>0) is distributed as
💡 Explanation:Linear transform of normal is normal.
- Q35hard
Sum of independent normal variables is
💡 Explanation:Normal family closed under independent sums.
- Q36Past Paper · PPSC/FPSC/CSSmedium
For X ~ N(10, 4), P(8 < X < 12) uses Z bounds
💡 Explanation:Z=(8−10)/2=−1 and (12−10)/2=+1.
- Q37Past Paper · PPSC/FPSC/CSSmedium
P(|Z| > 2) is approximately
💡 Explanation:Two tails beyond ±2: ≈0.0456 total.
- Q38Past Paper · PPSC/FPSC/CSSmedium
A test score X ~ N(70, 100). A score of 85 has Z-score
💡 Explanation:Z=(85−70)/10=1.5.
- Q39Past Paper · PPSC/FPSC/CSSmedium
Heights ~ N(170, 25). Proportion with height > 180 cm uses Z equal to
💡 Explanation:Z=(180−170)/5=2.
- Q40Past Paper · PPSC/FPSC/CSSmedium
Central Limit Theorem states that
💡 Explanation:CLT: X̄ₙ ≈ N(μ, σ²/n) for large n.
- Q41Past Paper · PPSC/FPSC/CSSmedium
For non-normal population with large n, distribution of X̄ is approximately
💡 Explanation:CLT applies regardless of population shape if n large.
- Q42Past Paper · PPSC/FPSC/CSSmedium
Standard error of the mean is
💡 Explanation:SE = σ/√n shrinks with sample size.
- Q43Past Paper · PPSC/FPSC/CSSeasy
If σ=20 and n=100, SE equals
💡 Explanation:SE = 20/√100 = 2.
- Q44hard
Normal probability plot is used to
💡 Explanation:QQ/plot checks normality assumption.
- Q45hard
Using normal approx for Bin(20, 0.4), μ=8, σ≈2.19. P(X≥10) with correction approximates P(Y>9.5)
💡 Explanation:P(X≥10)=P(X>9.5) with correction → P(Y>9.5).
- Q46medium
Skewed data with small n should
💡 Explanation:Small n or strong skew violates normal approx assumptions.
- Q47hard
The PDF of N(μ,σ²) has inflection points at
💡 Explanation:Second derivative zero at one standard deviation from mean.
- Q48medium
For continuous uniform on [a,b], P(X < c) for a < c < b equals
💡 Explanation:CDF of uniform is linear: (c−a)/(b−a).
- Q49hard
The 90th percentile of Z is approximately
💡 Explanation:P(Z≤1.28)≈0.90.
- Q50Past Paper · PPSC/FPSC/CSSmedium
For manufacturing: bolt diameter ~ N(10, 0.01). σ=0.1 mm. P(diameter < 9.8) uses Z=
💡 Explanation:Z=(9.8−10)/0.1=−2.