Respuesta :
Answer:
a) [tex]X \sim N(18,5)[/tex]
b) [tex]P(X>21)=P(\frac{X-\mu}{\sigma}>\frac{21-\mu}{\sigma})=P(Z<\frac{21-18}{5})=P(z>0.6)[/tex]
And we can find this probability with the complement rule and with the normal standard table or excel and we got:
[tex]P(z>0.6)=1-P(z<0.6)=1-0.726=0.274[/tex]
c) [tex]P(15<X<20)=P(\frac{15-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{21-\mu}{\sigma})=P(\frac{15-18}{5}<Z<\frac{20-18}{5})=P(-0.6<z<0.6)[/tex]
And we can find this probability with this difference and with the normal standard table or excel and we got:
[tex]P(-0.6<z<0.6)=P(z<0.6)-P(z<0.6)=0.726- 0.274=0.452[/tex]
d) For this case we can use the definition of z score, since we need 85% of the values in the middle of the distribution on the tails we need the remaining [tex] 1-0.85 = 0.15[/tex] and the value for [tex]\alpha/2 = 0.15/2= 0.075[/tex] and if we look that accumulates 0.075 of the area on the left of the normal standard distribution we got:
[tex]z = -1.440[/tex]
And using the z score formula we got:
[tex]-1.440 = \frac{X -18}{5}[/tex]
And solving for X we got : [tex] X= 18 -1.440*5 = 10.8[/tex]
And since the distribution is symmetrical we got:
[tex]1.440 = \frac{X -18}{5}[/tex]
And solving for X we got : [tex] X= 18 +1.440*5 = 25.2[/tex]
then we have 85% of the values between (10.8; 25.2)
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 duration of a particular type of criminal of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(18,5)[/tex]
Where [tex]\mu=18[/tex] and [tex]\sigma=5[/tex]
Part b
We are interested on this probability
[tex]P(X>21)[/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>21)=P(\frac{X-\mu}{\sigma}>\frac{21-\mu}{\sigma})=P(Z<\frac{21-18}{5})=P(z>0.6)[/tex]
And we can find this probability with the complement rule and with the normal standard table or excel and we got:
[tex]P(z>0.6)=1-P(z<0.6)=1-0.726=0.274[/tex]
Part c
[tex]P(15<X<20)=P(\frac{15-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{21-\mu}{\sigma})=P(\frac{15-18}{5}<Z<\frac{20-18}{5})=P(-0.6<z<0.6)[/tex]
And we can find this probability with this difference and with the normal standard table or excel and we got:
[tex]P(-0.6<z<0.6)=P(z<0.6)-P(z<0.6)=0.726- 0.274=0.452[/tex]
Part d
For this case we can use the definition of z score, since we need 85% of the values in the middle of the distribution on the tails we need the remaining [tex] 1-0.85 = 0.15[/tex] and the value for [tex]\alpha/2 = 0.15/2= 0.075[/tex] and if we look that accumulates 0.075 of the area on the left of the normal standard distribution we got:
[tex]z = -1.440[/tex]
And using the z score formula we got:
[tex]-1.440 = \frac{X -18}{5}[/tex]
And solving for X we got : [tex] X= 18 -1.440*5 = 10.8[/tex]
And since the distribution is symmetrical we got:
[tex]1.440 = \frac{X -18}{5}[/tex]
And solving for X we got : [tex] X= 18 +1.440*5 = 25.2[/tex]
then we have 85% of the values between (10.8; 25.2)
