VTU 3rd Sem BCS301 | Maths Module 4 |Central Limit Theorem Problem | Probability of Sample Mean |PYQ

Express VTU 4 All presents an important problem from Module 4 – Central Limit Theorem of VTU 3rd Semester Mathematics (BCS301). This question is frequently asked in VTU exams and is based on probability limits of sample mean using CLT. Question Use the Central Limit Theorem to evaluate P(50 less than or equal to X bar less than or equal to 56), where X bar represents the mean of a random sample of size 100 from an infinite population with mean μ = 53 and variance σ² = 400. Given A(1.5) = 0.4332. Step by Step Solution 1. Given Data Population mean μ = 53 Population variance σ² = 400 Standard deviation σ = 20 Sample size n = 100 Mean of X bar = μ = 53 Standard deviation of X bar = σ divided by square root of n = 20 divided by 10 = 2 2. Convert to Standard Normal Variable Lower limit: Z1 = (50 − 53) divided by 2 Z1 = −1.5 Upper limit: Z2 = (56 − 53) divided by 2 Z2 = 1.5 3. Probability Using Normal Table P(50 ≤ X bar ≤ 56) = P(−1.5 ≤ Z ≤ 1.5) From normal table, A(1.5) = 0.4332 So, P = 2 × 0.4332 P = 0.8664 Final Answer The required probability is 0.8664. Conclusion Using Central Limit Theorem, the probability that the sample mean lies between 50 and 56 is 86.64 percent. What You Will Learn in This Video Central Limit Theorem basics Standardization of sample mean Z value calculation Probability using normal table VTU exam solving strategy for Module 4 VTU 3rd Sem Maths Module 4 BCS301 Central Limit Theorem Probability of Sample Mean CLT solved problems VTU VTU Maths PYQ Sampling distribution of mean Express VTU 4 All #VTU #VTU3rdSem #VTUMaths #BCS301 #CentralLimitTheorem #CLT #Statistics #VTUPYQ #ExpressVTU4All