Question

A machine manufactures parts whose diameters vary according to the Normal distribution with mean 40.150 mm and standard deviation 0.003mm. An inspector measures a random sample of 4 parts. The probability that the average (x-bar) diameter of these 4 parts is less than 40.148 mm is about

Answer 1

The probability that the average (x-bar) diameter of these 4 parts is less than 40.148 mm is about 0.09176 or 9.2%

Explanation:

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".

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

`X \sim N(40.150,0.003)`

Where
`\mu=40.15` and
`\sigma=0.003`

They select a sample if size n=4, by the central limit theorem the distribution for the sample mean is given by:

`\bar X \sim N(\mu=40.150, (0.003)/(√(4)))`

We are interested on this probability

`P(\bar X<40.148)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(\bar x-\mu)/((\sigma)/(√(n)))`

If we use this formula we got this:

`P(\bar X<40.148)=P(Z<(40.148-40.15)/((0.003)/(√(4))))=P(Z<-1.33)=0.09176`

The probability that the average (x-bar) diameter of these 4 parts is less than 40.148 mm is about 0.09176 or 9.2%