Respuesta :
Answer:
[tex]P(8.6<X<9.8)=P(\frac{8.6-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{9.8-\mu}{\sigma})=P(\frac{8.6-9}{0.4}<Z<\frac{9.8-9}{0.4})=P(-1<z<2)[/tex]
And using the normal standard distirbution or excel we got:
[tex]P(-1<z<2)=P(z<2)-P(z<-1)= 0.97725-0.158655 = 0.7907[/tex]
Step-by-step explanation:
Let X the random variable that represent the resistance for resistors of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(9,0.4)[/tex]
Where [tex]\mu=9[/tex] and [tex]\sigma=0.4[/tex]
We are interested on this probability
[tex]P(8.6<X<9.8)[/tex]
And the z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(8.6<X<9.8)=P(\frac{8.6-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{9.8-\mu}{\sigma})=P(\frac{8.6-9}{0.4}<Z<\frac{9.8-9}{0.4})=P(-1<z<2)[/tex]
And using the normal standard distirbution or excel we got:
[tex]P(-1<z<2)=P(z<2)-P(z<-1)= 0.97725-0.158655 = 0.7907[/tex]
Answer:
P = 0.8185
Step-by-step explanation:
First we need to standardize the values 8.6 and 9.8 ohms using the following equation:
[tex]z=\frac{x-m}{s}[/tex]
Where m is the mean and s is the standard deviation, so 8.6 and 9.8 are equivalent to:
[tex]z=\frac{8.6-9}{0.4}=-1\\\\z=\frac{9.8-9}{0.4}=2[/tex]
Finally, using the normal distribution table, we can calculated the probability that a resistor has a resistance between 8.6 and 9.8 ohms as:
[tex]P(8.6<z<9.8)=P(-1<z<2)\\P(-1<z<2)=P(z<2)-P(z<-1)\\P(-1<z<2)=0.9772-0.1587\\P(-1<z<2)=0.8185[/tex]
