Question

A manufacturing facility produces rolls of tape. Among all of the rolls of tape produced, the mean length of the tape on a roll is 651.25 inches, and the standard deviation of the lengths is 0.73 inches. Suppose that we randomly select a sample of 78 rolls of tape. What is the probability that the sample mean length is less than 651.10 inches?

Answer 1

Answer:

$P(\bar X< 651.1) =P(\bar X< \frac{651.1-651.25}{\frac{0.73}{\sqrt{78}}}= -1.815$

And using the normal atandard table or excel we got:

$P(z$

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".  

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".

Solution to the problem

Let X the random variable that represent the length of the tape of a population, and for this case we know the following parameters

Where $\mu=651.25$ and $\sigma=0.73$

We sselect a sample size of n =78>30. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And we want to find this probability:

$P(\bar X< 651.1) =P(\bar X< \frac{651.1-651.25}{\frac{0.73}{\sqrt{78}}}= -1.815$

And using the normal atandard table or excel we got:

$P(z$