Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.4 and a mean diameter of 212 inches. If 80 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.1 inches? Round your answer to four decimal places.

Answer 1

Answer:

$P(211.9$  

And using the normal standard table or excel we can find this probability with this difference:

$P(-0.639$

Step-by-step explanation:

For this case we define the random variable X="metal shafts" produced in a manufacturing company, and we have the following properties given:

$E(X)= 212, \sigma = 1.4$

Previous concepts

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".  

Solution to the problem

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

$\bar X \sim N(\mu=212, \frac{1.4}{\sqrt{80}}= 0.1565)$

We are interested on this probability  

$P(211.9$  

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

$z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

If we use this formula we got this:

$P(211.9$  

And using the normal standard table or excel we can find this probability with this difference:

$P(-0.639$

Answer 2

Answer:

P (211.9 < Ẋ < 212.1) = P (211.9 -212/1.4√80 < Z < 212.1 – 212/1.4√80) P ( -0.639 < Z < 0.639)

We can find the probability with the difference, by using the normal standard table as shown,

P ( -0.639 < Z < 0.639) = P (Z < 0.639) -  P P (Z < - 0.639) = 0.7386 – 0.2614 = 0.4772

Step-by-step explanation:

For this example, X will be defined as a random variable  

X= Metal shafts produced by the manufacturing company, and the following properties is shown below

E (X) = 212, σ = 1.4

From the previous theorem,  

The Theorem known as central limit states that "if we have a standard deviation σ and a population with mean μ and take an amount of random samples that are large from the population with replacement, then the sample means of distribution will be approximately normally distributed. This is true irrespective of whether the source population is normal, provided the sample size is large enough

The Normal distribution is a probability distribution that is symmetric of the mean, shows that, data close to the mean occurs more frequently than data far from the mean.

A numerical measurement used in statistics of a relationship value to the mean of a group of values, measured in terms of standard deviation from the mean is known as the Z-score

Now,

Selecting a sample if size n=4,  

By applying the central limit theorem, the distribution of the sample mean is shown as,

Ẋ ~ N (µ = 212, 1.4/√80 = 0.1565)

The probability of area of interest will then be:

P (211.9 < Ẋ <212.1)

In solving this problem, the best approach will be using the normal standard distribution and z score which is,

z =ẋ - µ/σ √n

By applying this formula, we have at this:

P (211.9 < Ẋ <212.1) = P (211.9 -212/1.4√80 < Z < 212.1 – 212/1.4√80) P ( -0.639 < Z < 0.639)

Therefore, Using the normal standard excel or table we can find the probability of the difference which is now,

P ( -0.639 < Z < 0.639) = P (Z < 0.639) -  P P (Z < - 0.639) = 0.7386 – 0.2614 = 0.4772