Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205 inches. If 79 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 more than 0.3 inches

Answer 1

To find the mean diameter and standard deviation for the sample, use the provided formulas. Use the z-score formula to calculate the probability of the mean diameter differing from the population mean by more than 0.3 inches.

Step-by-step explanation:

To find the mean diameter and standard deviation for the sample, we need to use the formula:

Mean of sample (X-bar) = Population mean (μ)

Standard deviation of sample = Population standard deviation (σ) / √(sample size)

In this case, the population mean is 205 inches and the population standard deviation is 1.5 inches. The sample size is 79.

Substituting these values into the formulas, we get:

X-bar = 205 inches

Standard deviation of sample = 1.5 inches / √(79)

Then, to find the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches, we can use the z-score formula:

Z = (X - μ) / (σ / √n)

where X is the desired difference, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Substituting in the values given, we have:

Z = (0.3 - 0) / (1.5 / √79)

Calculating this, we find:

Z ≈ 1.4977

Using a z-table or calculator, we can find that the probability of a z-score greater than 1.4977 is approximately 0.067, or 6.7%.

Therefore, the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches is approximately 0.067, or 6.7%.

Answer 2

`P(205-0.3=204.7<\bar X<205+0.3=205.3)`

`z=(204.7-205)/((1.5)/(√(79)))=-1.778`

`z=(205.3-205)/((1.5)/(√(79)))=1.778`

So we can find this probability:

`P(-1.778<Z<1.778) = P(Z<1.778) -P(Z<-1.778) =0.962-0.0377= 0.9243`

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

`P = 1-0.9243 = 0.0757`

Step-by-step explanation:

Let X the random variable that represent the diamters of interest for this case, and for this case we know the following info

Where
`\mu=205` and
`\sigma=1.5`

We can begin finding this probability this probability

`P(205-0.3=204.7<\bar X<205+0.3=205.3)`

For this case they select a sample of n=79>30, so then we have enough evidence to use the central limit theorem and the distirbution for the sample mean can be approximated with:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

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)))`

And we can find the z scores for each limit and we got:

`z=(204.7-205)/((1.5)/(√(79)))=-1.778`

`z=(205.3-205)/((1.5)/(√(79)))=1.778`

So we can find this probability:

`P(-1.778<Z<1.778) = P(Z<1.778) -P(Z<-1.778) =0.962-0.0377= 0.9243`

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

`P = 1-0.9243 = 0.0757`