Question

The mean weight of loads of rock is 42.0 tons with a standard deviation of 12.0 tons. If 25 loads are chosen at random for a weight check, find the probability that the mean weight of those loads is less than 40.7 tons. Assume that the variable is normally distributed.

Answer 1

Answer:

$P(\bar X <40.7) =P(Z<-0.542) =0.294$

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".  

Solution to the problem

Let X the random variable that represent the weigths of rock from population, and for this case we know the distribution for X is given by:

$X \sim N(42,12)$  

Where $\mu=42$ and $\sigma=12$

The distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu= 42, \sigma_{\bar X}=\frac{\sigma}{\sqrt{n}}=\frac{12}{\sqrt{25}}=2.4)$

For this case we want to find this probability:

$P(\bar X <40.7)$

And for this we can use the z score given by:

$z = \frac{\bar x -\mu}{\sigma_{\bar X}}$

And if we use this formula we got:

$P(\bar X <40.7) =P(Z< \frac{40.7-42}{2.4}) = P(Z<-0.542)$

And if we use the normal standard table or excel we find the probability:

$P(\bar X <40.7) =P(Z<-0.542) =0.294$

Answer 2

The probability that the mean weight of those loads is less than 40.7 tons is  0.09%

z score is used to determine by how many standard deviations the raw score is above or below the mean.

The z score is given by:

$z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ x\ is\ raw\ score,\mu\ is\ mean, \sigma\ is\ standard\ deviatiom,n\ is \ sample\\size$

Given that μ = 42, σ = 12, n = 25. Hence for x < 40.7:

$z=\frac{40.7-42}{12/\sqrt{25} } =-3.12$

From the normal distribution table:

P(x < 40.7) = P(z < -3.12) = 0.0009 = 0.09%

The probability that the mean weight of those loads is less than 40.7 tons is  0.09%

Find out more on z score at: https://brainly.com/question/25638875