Respuesta :
Answer:
[tex]P(141.8<X<218)=P(\frac{141.8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{218-\mu}{\sigma})=P(\frac{141.8-173.6}{49.8}<Z<\frac{218-173.6}{49.8})=P(-0.639<z<0.892)[/tex]
And we can find this probability with this difference and using the normal standard table:
[tex]P(-0.639<z<0.892)=P(z<0.892)-P(z<-0.639)=0.814-0.261= 0.553[/tex]
Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women
Step-by-step explanation:
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(173.6,49.8)[/tex]
Where [tex]\mu=173.6[/tex] and [tex]\sigma=49.8[/tex]
We are interested on this probability
[tex]P(141.8<X<2188)[/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(141.8<X<218)=P(\frac{141.8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{218-\mu}{\sigma})=P(\frac{141.8-173.6}{49.8}<Z<\frac{218-173.6}{49.8})=P(-0.639<z<0.892)[/tex]
And we can find this probability with this difference and using the normal standard table:
[tex]P(-0.639<z<0.892)=P(z<0.892)-P(z<-0.639)=0.814-0.261= 0.553[/tex]
Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women
