Respuesta :
Answer:
a) [tex] P(492<X<512) = P(\frac{492-502}{10.54} <Z<\frac{512-502}{10.54}) = P(-0.949 < Z< 0.949)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-0.949 < Z< 0.949)= P(Z<0.949) -P(Z<-0.949) =0.8287-0.1713=0.6574 [/tex]
b) [tex] P(505<X<525) = P(\frac{505-515}{10.54} <Z<\frac{525-505}{10.54}) = P(-0.949 < Z< 1.898)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-0.949 < Z< 1.898)= P(Z<1.898) -P(Z<-0.949) =0.9712-0.1713=0.7998 [/tex]
c) [tex] P(484<X<504) = P(\frac{484-494}{10} <Z<\frac{504-494}{10}) = P(-1 < Z< 1)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-1 < Z< 1)= P(Z<1) -P(Z<-1) =0.8413-0.1587=0.6827 [/tex]
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 for critical reading of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(502,100)[/tex]
Where [tex]\mu=502[/tex] and [tex]\sigma=100[/tex]
We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)[/tex]
And for this case we want this probability:
[tex] P(502-10 < \bar X < 502+10)[/tex]
And for this case we can use the z score given by:
[tex] z= \frac{\bar X -\mu}{\sigma_{\bar x}}[/tex]
And if we use this formula we got:
[tex] P(492<X<512) = P(\frac{492-502}{10.54} <Z<\frac{512-502}{10.54}) = P(-0.949 < Z< 0.949)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-0.949 < Z< 0.949)= P(Z<0.949) -P(Z<-0.949) =0.8287-0.1713=0.6574 [/tex]
Part b
Let X the random variable that represent the scores for Math of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(515,100)[/tex]
Where [tex]\mu=515[/tex] and [tex]\sigma=100[/tex]
We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)[/tex]
And for this case we want this probability:
[tex] P(515-10 < \bar X < 515+10)[/tex]
And for this case we can use the z score given by:
[tex] z= \frac{\bar X -\mu}{\sigma_{\bar x}}[/tex]
And if we use this formula we got:
[tex] P(505<X<525) = P(\frac{505-515}{10.54} <Z<\frac{525-505}{10.54}) = P(-0.949 < Z< 1.898)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-0.949 < Z< 1.898)= P(Z<1.898) -P(Z<-0.949) =0.9712-0.1713=0.7998 [/tex]
Part c
Let X the random variable that represent the scores for Writing of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(494,100)[/tex]
Where [tex]\mu=494[/tex] and [tex]\sigma=100[/tex]
We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)[/tex]
And for this case we want this probability:
[tex] P(494-10 < \bar X < 494+10)[/tex]
And for this case we can use the z score given by:
[tex] z= \frac{\bar X -\mu}{\sigma_{\bar x}}[/tex]
And if we use this formula we got:
[tex] P(484<X<504) = P(\frac{484-494}{10} <Z<\frac{504-494}{10}) = P(-1 < Z< 1)[/tex]
And we can use excel or the normal standard table to find this probability:
[tex] P(-1 < Z< 1)= P(Z<1) -P(Z<-1) =0.8413-0.1587=0.6827 [/tex]
