Respuesta :
Answer:
a) [tex]P(X<16)=P(\frac{X-\mu}{\sigma}<\frac{16-\mu}{\sigma})=P(Z<\frac{16-21.1}{5.3})=P(z<-0.962)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.962)=0.168[/tex]
b) [tex]P(19<X<24)=P(\frac{19-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{24-\mu}{\sigma})=P(\frac{19-21.1}{5.3}<Z<\frac{24-21.1}{5.3})=P(-0.396<z<0.547)[/tex]
And we can find this probability with this difference:
[tex]P(-0.396<z<0.547)=P(z<0.547)-P(z<-0.396)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.396<z<0.547)=P(z<0.547)-P(z<-0.396)=0.708-0.346=0.362 [/tex]
c) [tex]P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.925)=1- P(Z<0.925) = 1-0.823 = 0.177[/tex]
d) We can consider unusual events values above or below 2 deviations from the mean
[tex] Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5[/tex]
A value below 10.5 can be consider as unusual
[tex] Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7[/tex]
A value abovr 31.7 can be consider as unusual
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 scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(21.1,5.3)[/tex]
Where [tex]\mu=21.1[/tex] and [tex]\sigma=5.3[/tex]
We are interested on this probability
[tex]P(X<16)[/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<16)=P(\frac{X-\mu}{\sigma}<\frac{16-\mu}{\sigma})=P(Z<\frac{16-21.1}{5.3})=P(z<-0.962)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.962)=0.168[/tex]
Part b
[tex]P(19<X<24)=P(\frac{19-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{24-\mu}{\sigma})=P(\frac{19-21.1}{5.3}<Z<\frac{24-21.1}{5.3})=P(-0.396<z<0.547)[/tex]
And we can find this probability with this difference:
[tex]P(-0.396<z<0.547)=P(z<0.547)-P(z<-0.396)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.396<z<0.547)=P(z<0.547)-P(z<-0.396)=0.708-0.346=0.362 [/tex]
Part c
[tex]P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.925)=1- P(Z<0.925) = 1-0.823 = 0.177[/tex]
Part d
We can consider unusual events values above or below 2 deviations from the mean
[tex] Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5[/tex]
A value below 10.5 can be consider as unusual
[tex] Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7[/tex]
A value abovr 31.7 can be consider as unusual
