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

Answer:

$P(205-0.3=204.7$

$z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778$

$z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778$

So we can find this probability:

$P(-1.778$

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

$P(205-0.3=204.7$

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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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

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

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

$z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778$

$z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778$

So we can find this probability:

$P(-1.778$

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:

$P = 1-0.9243 = 0.0757$