The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 7/8 inches and a standard deviation of 1/16 inch. Assume independence. a. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. b. What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? c. What is the probability that the door does not fit in the casing?

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!!!

Explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

if we substitute the value of the number of lattice into the first equation, we arrive at