Respuesta :
Answer:
[tex] P(X<95) =P(Z<\frac{95-100}{10}) =P(Z<-0.5) = 0.309[/tex]
And for the other case if we use the z score and the complement rule we have:
[tex] P(X>105) =P(Z>\frac{105-100}{10}) =1-P(Z<0.5) =1-0.691=0.309[/tex]
And we can find the probability of interest like this:
[tex]P(X<95 \cup X>105)=P(X<95) +P(X>105) -P(X<95 \cap X>105)[/tex]And replacing we got:
[tex]P(X<95 \cup X>105)=0.309+0.309-0= 0.618[/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".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
We assume that the distribution is [tex]X \sim N(100,10)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=10[/tex]
We are interested on this probability
[tex]P(X<95 \cup X>105)[/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 find the individual probabilities we got:
[tex] P(X<95) =P(Z<\frac{95-100}{10}) =P(Z<-0.5) = 0.309[/tex]
And for the other case if we use the z score and the complement rule we have:
[tex] P(X>105) =P(Z>\frac{105-100}{10}) =1-P(Z<0.5) =1-0.691=0.309[/tex]
And we can find the probability of interest like this:
[tex]P(X<95 \cup X>105)=P(X<95) +P(X>105) -P(X<95 \cap X>105)[/tex]And replacing we got:
[tex]P(X<95 \cup X>105)=0.309+0.309-0= 0.618[/tex]
