Question

A machine at Katz Steel Corporation makes 5-inch-long nails. The probability distribution of the lengths of these nails is approximately normal with a mean of 5 inches and a standard deviation of 0.12 inch. The quality control inspector takes a sample of 36 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than 4.94 inches or greater than 5.06 inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 36 nails, the inspector will conclude that the machine needs an adjustment?

Answer 1

69

Explanation:

70-1=69

Answer 2

The answer for this case would be 0.0027.

Explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable who represents the lengths the nails produced by a machine at Katz Steel Corporation. We know from the problem that the distribution for the random variable X is given by:

`X\sim N(\mu =5,\sigma =0.12)`

The quality control inspector takes a sample of n=36 nails. That represent the sample size.

And we want to calculate the probability that the mean of this sample is either less than 4.94 inches or greater than 5.06 inches. So we want to find:

`P(\bar X <4.94) or P(\bar x>5.06)`

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the individual probabilities like this:

`P(\bar X <4.94)=P(Z<(4.94-5)/((0.12)/(√(36)))=-3)`

And using a calculator, excel ir the normal standard table we have that:

`P(Z<-3)=0.00135`

And for the other individual probability we have:

`P(\bar X >5.06)=P(Z>(5.06-5)/((0.12)/(√(36)))=3)`

`P(Z>3)=1-P(Z<3)= 1-0.99865=0.00135`

And then we can find the probability required on this way:

`P(\bar X <4.94) or P(\bar X>5.06)=P(\bar X <4.94)+P(\bar x>5.06)-[P(\bar X<4.94)andP(\bar X >5.06)]`

`=P(\bar X <4.94)+P(\bar x>5.06)-0= 0.00135+0.00135=0.0027`

What is the probability that based on a sample of 36 nails, the inspector will conclude that the machine needs an adjustment?

The answer for this case would be 0.0027.