Question

Find the specified areas for a normal density. (a) The area above 200 on a \(N(120,40)\) distribution (b) The area below 49.5 on a \(N(50,0.2)\) distribution (c) The area between 0.8 and 1.5 on a \(N(1,0.3)\) distribution

Short answer

(a) The area above 200 in a \(N(120,40)\) distribution is about 0.0228.\n(b) The area below 49.5 in a \(N(50,0.2)\) distribution is approximately 0.0062.\n(c) The area between 0.8 and 1.5 in a \(N(1,0.3)\) distribution is roughly 0.7011.

Step-by-step solution

Solve for Part (a)

First, we need to calculate the Z-score for the value 200 on a \(N(120,40)\) distribution using the formula: Z = (X - μ) / σ Then, substitute the values into the formula: Z = (200 - 120) / 40 = 2 Since we need to find the area above this Z-score, we find the area corresponding to Z = 2 from a standard normal distribution table, which is 0.9772. But this is the area below Z = 2, so to find the area above Z = 2, we subtract this value from 1: Area = 1 - 0.9772 = 0.0228

Solve for Part (b)

Once again we use the formula for the Z-score, this time for the value 49.5 on a \(N(50,0.2)\) distribution: Z = (49.5 - 50) / 0.2 = -2.5 We want to find the area below this Z-score, so we look up the value for Z = -2.5 in a standard normal distribution table, and we get Area = 0.0062

Solve for Part (c)

This part asks for the area between two values under a \(N(1,0.3)\) distribution. First, we calculate the Z-scores for these two values:Z₁ = (0.8 - 1) / 0.3 = -0.67 (approximately) and Z₂ = (1.5 - 1) / 0.3 = 1.67 (approximately) We need to find the areas below these Z-scores from a standard normal distribution table, Area₁ = 0.2514 and Area₂ = 0.9525. But we want the area between these Z-scores, so we subtract Area₁ from Area₂: Area = Area₂ - Area₁ = 0.9525 - 0.2514 = 0.7011