Respuesta :
Answer:
a) [tex]P(X<0.066)=P(\frac{X-\mu}{\sigma}<\frac{0.066-\mu}{\sigma})=P(Z<\frac{0.066-0.1}{0.02})=P(Z<-1.7)[/tex]
And we can find this probability using the normal standard distribution or excel:
[tex]P(Z<-1.7)=0.045[/tex]
b) [tex]P(0.07<X<0.122)=P(\frac{0.07-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{0.122-\mu}{\sigma})=P(\frac{0.07-0.1}{0.02}<Z<\frac{0.122-0.1}{0.02})=P(-1.5<Z<1.1)[/tex]
[tex]P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)=0.864-0.066=0.798[/tex]
c) [tex] P(0.07<X<0.122)=0.995[/tex]
And for this case we can use the z score formula given by:
[tex] z= \frac{x -\mu}{\sigma}[/tex]
We find a value on the normal standard distribution that accumulates 0.0025 of the both tails and the values for this case are [tex] z= \pm 2.81[/tex]
And we can use the z score like this:
[tex] -2.81 = \frac{0.07-0.1}{\sigma}[/tex]
And solving for the deviation we got:
[tex] \sigma= \frac{0.07-0.1}{-2.81}= 0.0107[/tex]
[tex] 2.81 = \frac{0.122-0.1}{\sigma}[/tex]
And solving for the deviation we got:
[tex] \sigma= \frac{0.122-0.1}{2.81}= 0.00783[/tex]
So then we can conclude that the standard deviation needs to be between [tex] 0.00783 \leq \sigma \leq 0.0107[/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 diameter of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(0.1,0.02)[/tex]
Where [tex]\mu=0.1[/tex] and [tex]\sigma=0.02[/tex]
We are interested on this probability
[tex]P(X<0.066)[/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(X<0.066)=P(\frac{X-\mu}{\sigma}<\frac{0.066-\mu}{\sigma})=P(Z<\frac{0.066-0.1}{0.02})=P(Z<-1.7)[/tex]
And we can find this probability using the normal standard distribution or excel:
[tex]P(Z<-1.7)=0.045[/tex]
Part b
[tex]P(0.07<X<0.122)=P(\frac{0.07-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{0.122-\mu}{\sigma})=P(\frac{0.07-0.1}{0.02}<Z<\frac{0.122-0.1}{0.02})=P(-1.5<Z<1.1)[/tex]
And we can find this probability with this difference:
[tex]P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)=0.864-0.066=0.798[/tex]
Part c
For this case w eneed this condition:
[tex] P(0.07<X<0.122)=0.995[/tex]
And for this case we can use the z score formula given by:
[tex] z= \frac{x -\mu}{\sigma}[/tex]
We find a value on the normal standard distribution that accumulates 0.0025 of the both tails and the values for this case are [tex] z= \pm 2.81[/tex]
And we can use the z score like this:
[tex] -2.81 = \frac{0.07-0.1}{\sigma}[/tex]
And solving for the deviation we got:
[tex] \sigma= \frac{0.07-0.1}{-2.81}= 0.0107[/tex]
[tex] 2.81 = \frac{0.122-0.1}{\sigma}[/tex]
And solving for the deviation we got:
[tex] \sigma= \frac{0.122-0.1}{2.81}= 0.00783[/tex]
So then we can conclude that the standard deviation needs to be between [tex] 0.00783 \leq \sigma \leq 0.0107[/tex]
