Respuesta :
Answer:
a)[tex]z=\frac{1000-1065}{56}=-1.16[/tex]
And this value means that the weight X=1000 pounds it's 1.6 deviations below the true mean of 1065 pounds
b) [tex]z=\frac{1250-1065}{56}=3.304[/tex]
And this value means that the weight X=1250 pounds it's 3.304 deviations above the true mean of 1065 pounds.
So we can cocnlude on this case that the weight of 1250 would me more unusual since represent a value with a more distance respect to the mean than the weight of 1000 pounds.
Step-by-step explanation:
1) 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".
2) Part a
Let X the random variable that represent the weights of a breed of yearling cattle of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1065,56)[/tex]
Where [tex]\mu=1065[/tex] and [tex]\sigma=56[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we find the Z score for the value of X=1000 we got:
[tex]z=\frac{1000-1065}{56}=-1.16[/tex]
And this value means that the weight X=1000 pounds it's 1.6 deviations below the true mean of 1065 pounds
3) Part b
For this case if we find the z score for the value of 1250 we got:
[tex]z=\frac{1250-1065}{56}=3.304[/tex]
And this value means that the weight X=1250 pounds it's 3.304 deviations above the true mean of 1065 pounds.
So we can cocnlude on this case that the weight of 1250 would me more unusual since represent a value with a more distance respect to the mean than the weight of 1000 pounds.
The more a value is away from mean of normal distribution, the more unusual it is. The values of needed figures for the given context are:
- X = 1000 is 1.16 standard deviations away(in left side as sign of Z is -ve) from the mean.
- X = 1000 is more near to mean than X = 1250, thus, X = 1250 is more unusual weight.
How to identify which value is more unusual in a normal distribution?
A normal distribution has the most likely value as its mean. The more a value goes away from that mean, the more unlikely it becomes for the considered random variable to assume that value.
For the given case, let the weights of all such animals be tracked by random variable X.
Then by the given data, we have:
[tex]X \sim N(1065, 56)[/tex]
Where mean of X = 1065, its standard deviation = 56
Evaluating the results for given cases:
- a) X = 1000, suppose is Z standard deviations away from its mean,then
[tex]\mu + Z \sigma = 1000\\\\Z= \dfrac{1000 - \mu}{\sigma} = \dfrac{1000 - 1065}{56} \approx -1.16[/tex]
Thus, X = 1000 is 1.16 standard deviations away(in left side as sign of Z is -ve) from the mean.
- b) X = 1000 is 1.16 standard deviations to the left of the mean of X
and for X = 1250, it is [tex]\dfrac{1250 - 1065}{56} \approx 3.3\\[/tex] (thus, x = 1250 is approx 3.3 standard deviations to the right of the mean)
Thus, X = 1000 is more near to mean than X = 1250, thus, X = 1250 is more unusual weight.
Also, know the fact that [tex]Z = \dfrac{X - \mu}{\sigma}[/tex]normal distribution is called standard normal variate. Its mean is 0 and variance is 1 and its like so much useful. The value we get for z for corresponding value of X is called z score.
The values of needed figures for the given context are:
- X = 1000 is 1.16 standard deviations away(in left side as sign of Z is -ve) from the mean.
- X = 1000 is more near to mean than X = 1250, thus, X = 1250 is more unusual weight.
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
