Answer:
a) Mean = 0.125 inch
Standard deviation = 0.13975 inch
b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25) = 0.18673
c) Probability that the door does not fit in the casing = P(X < 0) = 0.18673
Step-by-step explanation:
Let the distribution of the width of the casing be X₁ (μ₁, σ₁²)
Let the distribution of the width of the door be X₂ (μ₂, σ₂²)
The distribution of the difference between the width of the casing and the width of the door = X = X₁ - X₂
when two independent normal distributions are combined in any manner, the resulting distribution is also a normal distribution with
Mean = Σλᵢμᵢ
λᵢ = coefficient of each disteibution in the manner that they are combined
μᵢ = Mean of each distribution
Combined variance = σ² = Σλᵢ²σᵢ²
λ₁ = 1, λ₂ = -1
μ₁ = 24 inches
μ₂ = 23 7/8 inches = 23.875 inches
σ₁² = (1/8)² = (1/64) = 0.015625
σ₂ ² = (1/16)² = (1/256) = 0.00390625
Combined mean = μ = 24 - 23.875 = 0.125 inch
Combined variance = σ² = (1² × 0.015625) + [(-1)² × 0.00390625] = 0.01953125
Standard deviation = √(Variance) = √(0.01953125) = 0.1397542486 = 0.13975 inch
b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25)
This is a normal distribution problem
Mean = μ = 0.125 inch
Standard deviation = σ = 0.13975 inch
We first normalize/standardize 0.25 inch
The standardized score of any value is that value minus the mean divided by the standard deviation.
z = (x - μ)/σ = (0.25 - 0.125)/0.13975 = 0.89
P(X > 0.25) = P(z > 0.89)
Checking the tables
P(x > 0.25) = P(z > 0.89) = 1 - P(z ≤ 0.89) = 1 - 0.81327 = 0.18673
c) Probability that the door does not fit in the casing
If X₂ > X₁, X < 0
P(X < 0)
We first normalize/standardize 0 inch
z = (x - μ)/σ = (0 - 0.125)/0.13975 = -0.89
P(X < 0) = P(z < -0.89)
Checking the tables
P(X < 0) = P(z < -0.89) = 0.18673
Hope this Helps!!!
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