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Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15, 30, and 20 min, respectively, and the standard deviations are 2, 1, and 1.6 min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component? (Round your answer to four decimal places.)

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Answer:

0.0359

Step-by-step explanation:

Data provided:

mean values of three independent times are 15, 30, and 20 minutes

the standard deviations are 2, 1, and 1.6 minutes

Now,

New Mean = 15 + 30 + 25 = 65

Variance = ( standard deviation )²

or

Variance = 2² + 1² + 1.6² = 7.56

therefore,

Standard deviation = √variance

or

Standard deviation =  2.75

Thus,

Z-value = [tex]\frac{\textup{60 - 65}}{\textup{2.75}}[/tex]

or

Z-value = - 1.81

from the Z-table

the Probability of Z ≤ -1.81 = 0.0359

The probability that it takes at most 1 hour of machining time to produce a randomly selected component is 0.0344 approx.

What is the distribution of random variable which is sum of normal distributions?

Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:

[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]

And if

[tex]X = X_1 + X_2 + \cdots + X_n[/tex]

Then, its distribution is given as:

[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]

For the given case, if we take:

  • [tex]X_1 =[/tex]Time taken by first operation
  • [tex]X_2 =[/tex]second operation
  • [tex]X_3 =[/tex]operation

Then, by the given data we get:

[tex]X_1 \sim N(15, 2^2)\\X_2 \sim N(30,1^2)\\X_3 \sim N(20, 1.6^2)[/tex]

(since standard deviation's square is variance)

Then, Let X denotes the total time taken by all 3 operations to manufacture the product, then:

X = [tex]X_1 + X_2 + X_3[/tex] = Total time taken for manufacturing

Then, the needed probability is:

[tex]P(\text{total time is at max 1 hour = 60 minutes}) = P(X \leq 60)[/tex] (as all other random variables are doing calculating in minutes).

Since it is given that all three operations are independent, thus, we get the distribution of X is given as:

[tex]X \sim N(15 + 30 + 20, 2^2 + 1^2 + 1.6^2)\\\\X \sim N(65, 7.56)[/tex]

Here

[tex]\mu = 65,\\\sigma^2 = 7.56[/tex]

Now using the conversion to standard normal variate and the z-tables to calculate the needed probability, we get:

[tex]Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 65}{\sqrt{7.56}} \approx \dfrac{X - 65}{2.749}[/tex] (positive root since standard deviation is a non negative quantity)

The needed probability becomes

[tex]P(X \leq 60) = P(Z \leq \dfrac{60 - 65}{2.745}) \approx P(Z \leq -1.82)[/tex]

From the z-tables, we get the p value for Z = -1.82 is p = 0.0344

(p value for Z = z shows [tex]P(Z \leq z)[/tex] )

Thus, we get:

[tex]P(X \leq 60) \approx P(Z \leq -1.82) \approx 0.0344\\P(X \leq 60) \approx 0.0344[/tex]

The probability that it takes at most 1 hour of machining time to produce a randomly selected component is 0.0344 approx.

Learn more about standard normal distribution here:

https://brainly.com/question/10984889

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