1. Test scores are normal with mean 70 and SD 10. What is P(score > 85)?
z = (85−70)/10 = 1.5. From standard normal, P(Z>1.5) ≈ 0.0668.
z = 1.5 ⇒ P(Z>1.5)≈0.0668
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Aptitude & Reasoning · Question Set
Probability & Statistics (Basics) interview questions for placements and exams.
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Aptitude & Reasoning
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z = (85−70)/10 = 1.5. From standard normal, P(Z>1.5) ≈ 0.0668.
z = 1.5 ⇒ P(Z>1.5)≈0.0668
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Strength is by absolute value; negative sign indicates inverse relation. |−0.85| is largest among choices.
|r| largest ⇒ strongest
P(R then B) = (26/52)×(26/51) = (1/2)×(26/51) = 13/51.
(26/52)*(26/51)=13/51
Sorted data: 1,3,5,7,9,11. With even n, median is average of middle two: (5+7)/2 = 6.
Median = (x3 + x4)/2 = (5+7)/2
Mode is the most frequent value; 4 appears three times (maximum).
freq(4)=3 (max)
μ=2.5. Variance = average squared deviation: [(1−2.5)^2 + (2−2.5)^2 + (3−2.5)^2 + (4−2.5)^2]/4 = 5/4 = 1.25.
σ² = (2.25+0.25+0.25+2.25)/4 = 1.25
Binomial: C(5,3)(1/2)^5 = 10/32 = 5/16.
C(5,3)/2^5 = 10/32
P(X=3) = e^{−2} 2^3 / 3! = e^{−2}·8/6 ≈ 0.180.
P(3) = e^{-2} * 8/6Empirical rule: ~68% within 1σ, ~95% within 2σ, ~99.7% within 3σ.
≈95% within ±2σ
Systematic sampling chooses every k-th unit after a random start.
Pick start r∈{1..k}, then r, r+k, r+2k…E[X] = (1+2+3+4+5+6)/6 = 21/6 = 3.5.
E[X]=Σx·(1/6)=3.5
E[X]=3.5, E[X²]=(1²+…+6²)/6=91/6. Var = E[X²]−(E[X])² = 91/6 − 12.25 = 35/12 ≈ 2.917.
Var = 91/6 − (7/2)² = 35/12
Spades:13, Aces:4, overlap Ace of Spades:1. Favorable = 13+4−1 = 16 → 16/52 = 4/13.
(13+4−1)/52 = 16/52
Mean is sum divided by count: (2+4+6+8+10)/5 = 30/5 = 6.
(Σxi)/n = 30/5 = 6
P(A∪B) = P(A)+P(B)−P(A∩B) = 0.4+0.5−0.2 = 0.7.
P(A∪B)=0.4+0.5−0.2
Use complement: 1 − P(no six) = 1 − (5/6)^2 = 1 − 25/36 = 11/36.
1 − (5/6)^2 = 11/36
Bayes: P(D|+) = 0.01×0.99 / [0.01×0.99 + 0.99×0.05] = 0.0099 / 0.0594 ≈ 0.167.
0.0099/(0.0099+0.0495)≈0.167