Suppose that a random sample of size 25 is to be selected from a population with mean 41 and standard deviation 9. What is the approximate probability that X will be more than 0.5 away from the population mean? a) 0.7812b) 0.2188c) 0.4188d) 0.0443e) 0.4376f) None of the above

Answer:The probability that X will be more than 0.5 away fro the population mean is 0.7812 ⇒ answer aStep-by-step explanation:* Let us explain how to solve the problem- The sample size n is 25- The mean of the population μ is 41- The standard deviation σ is 9- We need to find the approximate probability that X will be more than  0.5 away from the population mean∵ z-score = (X - μ)/(σ/√n)∵ X - μ = 0.5- That means X > 41 + 0.5, OR X < 41 - 0.5- Then we need to find P(X > 41.5) and P(X < 40.5)∵ z = [tex]\frac{41.5-41}{\frac{9}{\sqrt{25}}}[/tex] = 0.2778∵ z = [tex]\frac{40.5-41}{\frac{9}{\sqrt{25}}}[/tex] = -0.2778- Let us use the normal distribution table to find the corresponding   area to z-score∵ P(z < 0.2778) = 0.6116∵ P(z > -0.2778) = 0.3928- P(40.5 < X < 41.5) = P( -0.2778 < z < 0.2778)∴ The probability of X to be with in 0.5 = 0.6116 - 0.3928∴ The probability of X to be with in 0.5 = 0.2188∵ The probability of X to be more than 0.5 away from the population    mean = 1 - 0.2188∴ P(X > 41.5) OR P(X < 40.5) = 0.7812* The probability that X will be more than 0.5 away fro the population   mean is 0.7812