Nuts and bolts are made separately and paired at random. The nut diameters (in mm) are normally distributed with mean 10 and variance .02. The bolt diameters (in mm) are also normally distributed with mean 9.5 and variance 0.02. Find the probability that a bolt is too large for its nut. In other words, find the probability that for a randomly selected nut and bolt, [size of nut − size of bolt] ≤ 0. (Alternatively, you could subtract in the ‘other’ direction and look for the probability that [size of bolt − size of nut] ≥ 0). Draw the distribution of the combination.

- 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!!!

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step-by-step explanation:

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