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A manufacturing process is designed to produce bolts with a 0.5-inch diameter. Once each day, a random sample of 36 bolts is selected and the bolt diameters recorded. If the resulting sample mean is less than 0.490 inches or greater than 0.510 inches, the process is shut down for adjustment. The standard deviation for diameter is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily

Respuesta :

Given Information:  

Population mean = μ = 0.5  

Population standard deviation = σ = 0.02

Sample size n = 36

Required Information:  

Probability that the manufacturing line will be shut down unnecessarily = ?  

Answer:  

P = 0.9975

Explanation:  

The sample mean will be same as population mean μs = 0.5

The sample standard deviation is given by

σs = σ/√n

σs = 0.02/√36

σs = 0.0033

Now we can calculate the probability that the process will shut down for adjustment.

P(x < 0.490) = (x - μs)/σs

P(x < 0.490) = (0.490 - 0.50)/0.0033

P(x < 0.490) = -3.03

The corresponding z-score from the z table is 0.00122

P(x > 0.510) = (x - μs)/σs

P(x > 0.510) = (0.510 - 0.50)/0.0033

P(x > 0.510) = 3.03

As you can see the distribution is symmetric, therefore,

P( 0.510 < x < 0.490) = 2*0.00122

P( 0.510 < x < 0.490) = 0.00244

Notice that it was asked in the question to find the probability that the manufacturing line will be shut down unnecessarily, which means

P = 1 - P( 0.510 < x < 0.490)

P = 1 - 0.00244

P = 0.9975

Therefore, the probability that the manufacturing line will be shut down unnecessarily is 0.9975.

Using the normal distribution and the central limit theorem, it is found that there is a 0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

  • The mean is of 0.5 inches, hence [tex]\mu = 0.5[/tex].
  • The standard deviation is of 0.02 inches, hence [tex]\sigma = 0.02[/tex].
  • A sample of 36 bolts is taken, hence [tex]n = 36, s = \frac{0.02}{\sqrt{36}} = 0.0033[/tex]

The manufacturing line will be shut down unnecessarily if the sample mean is less than 0.490 inches or greater than 0.510 inches.

  • The normal distribution is symmetric, which means that these probabilities are equal, hence, we find one of them and multiply by 2.
  • The probability that the resulting sample mean is less than 0.490 inches is the p-value of Z when X = 0.49, then:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.49 - 0.5}{0.0033}[/tex]

[tex]Z = -3[/tex]

[tex]Z = -3[/tex] has a p-value of 0.0013.

2 x 0.0013 = 0.0026.

0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

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