Q86. X α 1 0 −3 Let the mean and the standard deviation of the probability distribution be μ and σ, P(X) 31 K 16 41 respectively. If σ −μ = 2, then σ + μ is equal to________ JEE Main 2024 (05 Apr Shift 2) JEE Main Previous Year Paper
What This Question Tests
This question requires calculating the mean and standard deviation of a discrete probability distribution using the given formulas. It involves solving for an unknown probability and then applying the definitions of mean and variance.
Concepts Tested
Formulas Used
Sum of P(X) = 1
μ = Σ X P(X)
σ^2 = Σ X^2 P(X) - μ^2
σ = sqrt(σ^2)
📚 NCERT Sections This Tests
3.10 — In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 · Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
3.26 — The Decomposition Of Hydrocarbon Follows The Equation
Chemistry Class 11 · Chapter 3
3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26
📋 Question Details
- Chapter
- Statistics
- Topic
- Mean and standard deviation of discrete probability distribution
- Year
- 2024
- Shift
- 05 Apr Shift 2
- Q Number
- Q86
- Type
- Numerical
- NCERT Ref
- Class 12 Mathematics Ch 13: Probability
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