Respuesta :
Answer:
a) [tex]P(140<X<181)=P(\frac{140-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{181-\mu}{\sigma})=P(\frac{140-139}{28.1}<Z<\frac{181-139}{28.1})=P(0.036<z<1.495)[/tex]
[tex]P(0.036<z<1.495)=P(z<1.495)-P(z<0.036)[/tex]
[tex]P(0.036<z<1.495)=P(z<1.495)-P(z<0.036)=0.933-0.514=0.419[/tex]
b) [tex]P(140< \bar X <181)= P(Z<8.45)-P(Z<0.201) = 1-0.580=0.420[/tex]
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".
Part a
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(139,28.1)[/tex]
Where [tex]\mu=139[/tex] and [tex]\sigma=28.1[/tex]
We are interested on this probability
[tex]P(140<X<181)[/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 apply this formula to our probability we got this:
[tex]P(140<X<181)=P(\frac{140-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{181-\mu}{\sigma})=P(\frac{140-139}{28.1}<Z<\frac{181-139}{28.1})=P(0.036<z<1.495)[/tex]
And we can find this probability like this:
[tex]P(0.036<z<1.495)=P(z<1.495)-P(z<0.036)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(0.036<z<1.495)=P(z<1.495)-P(z<0.036)=0.933-0.514=0.419[/tex]
Part b
For this case we select a sample size of n =32. Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu=139, \frac{\sigma}{\sqrt{n}}=\frac{28.1}{\sqrt{32}}=4.97)[/tex]
And the new z score would be:
[tex] Z=\frac{\bar X -\mu}{\sigma_{\bar x}}[/tex]
[tex]P(140< \bar X <181)=P(\frac{140-139}{4.97}<Z<\frac{181-139}{4.97})[/tex]
[tex]P(140< \bar X <181)= P(Z<8.45)-P(Z<0.201) = 1-0.580=0.420[/tex]
