Answer:
- The probability that a bolt is too large for its nut = 0.00621
- The image of the drawing of this combined distribution is shown in the attached file to this solution.
Step-by-step explanation:
When independent, normal distributions are combined, the combined mean and combined variance are given through the relation
Combined mean = Σ λᵢμᵢ
(summing all of the distributions in the manner that they are combined)
Combined variance = Σ λᵢ²σᵢ²
(summing all of the distributions in the manner that they are combined)
For this question, the first distribution is the size of a nut, with mean 10 mm and a variance of 0.02
Second distribution is the size of a bolt, with mean 9.5 mm and a variance of 0.02.
The combined distribution is [size of nut − size of bolt]
Hence, λ₁ = 1, λ₂ = -1
μ₁ = 10 mm
μ₂ = 9.5 mm
σ₁² = 0.02
σ₂² = 0.02
Combined Mean = (1×10) + (-1×9.5) = 0.5 mm
Combined Variance = [(1)² × 0.02] + [(-1)² × 0.02] = 0.04
So, the combined distribution is also a normal distribution with a Mean of 0.5 mm and a variance of 0.04.
Standard deviation = √variance = √0.04 = 0.2 mm
The probability that a bolt is too large for its nut, [size of nut − size of bolt] ≤ 0, P(X ≤ 0)
To obtain this required probability, we first normalize/standardize 0 mm
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0 - 0.5)/0.2 = - 2.50
To determine the required probability
P(x ≤ 0) = P(z ≤ -2.50)
We'll use data from the normal probability table for these probabilities
P(x ≤ 0) = P(z ≤ -2.50) = 0.00621
The image of the drawing of this combined distribution is shown in the attached file to this solution.
Hope this Helps!!!
