Question

On average, the parts from a supplier have a mean of 35.8 inches and a standard deviation of 2.4 inches. Find the probability that a randomly selected part from this supplier will have a value between 28.6 and 43.0 inches. Is this consistent with the Empirical Rule of 68%-95%-99.7%?

Answer 1

Answer:

$P(28.6$

And on this case since we are within 3 deviations from the mean the result obtained using the z score agrees with the empirical rule.

Step-by-step explanation:

Previous concepts

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 who represent the parts from a supplier of a population, and for this case we know the distribution for X is given by:

$X \sim N(35.8,2.4)$  

Where $\mu=35.8$ and $\sigma=2.4$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

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

Find the probability that a randomly selected part from this supplier will have a value between 28.6 and 43.0 inches

$P(28.6$

And that correspond with the 99.73% of the data.

The empirical rule, also referred to as "the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)". The empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

And on this case since we are within 3 deviations the result obtained using the z score agrees with the empirical rule.