Respuesta :
Answer:
If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.
a) [tex]P(75<X<105)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(75<X<105) =P(\frac{75-90}{15} <Z<\frac{105-90}{15}) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683[/tex]
b) [tex]P(60<X<120)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(60<X<120) =P(\frac{60-90}{15} <Z<\frac{120-90}{15}) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955[/tex]
c) [tex]P(45<X<135)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(45<X<135) =P(\frac{45-90}{15} <Z<\frac{135-90}{15}) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997[/tex]
If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.
Step-by-step explanation:
Previous concepts
If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.
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 of interest
We assume for this case that [tex]\mu=90[/tex] and [tex]\sigma=15[/tex]
We are interested on this probability
[tex]P(75<X<105)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(75<X<105) =P(\frac{75-90}{15} <Z<\frac{105-90}{15}) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683[/tex]
Part b
[tex]P(60<X<120)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(60<X<120) =P(\frac{60-90}{15} <Z<\frac{120-90}{15}) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955[/tex]
Part c
[tex]P(45<X<135)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
And if we use it we got:
[tex] P(45<X<135) =P(\frac{45-90}{15} <Z<\frac{135-90}{15}) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997[/tex]
If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.
