Respuesta :
Answer:
[tex]\bar X \sim N(\mu. \frac{\sigma}{\sqrt{n}})[/tex]
And we want to find the following probability:
[tex] P(\bar X >619)[/tex]
And we can use the z score formula given by:
[tex] z=\frac{\bar x -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z=\frac{619-606}{\frac{62}{\sqrt{99}}}= 2.086[/tex]
And we can use the normal standard distirbution and the complement rule to find the probability:
[tex] P(z>2.086)=1 -P(z<2.086)= 1-0.982= 0.018[/tex]
Step-by-step explanation:
For this problem we have the following parameters given:
[tex]\mu = 606[/tex] represent the mean
[tex]\sigma = 62[/tex] represent the true deviation
[tex] n= 99[/tex] represent the sample size
For this case since the sample size is >30 we can use the central limit theorem and we can u se the following distribution for the sample mean
[tex]\bar X \sim N(\mu. \frac{\sigma}{\sqrt{n}})[/tex]
And we want to find the following probability:
[tex] P(\bar X >619)[/tex]
And we can use the z score formula given by:
[tex] z=\frac{\bar x -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z=\frac{619-606}{\frac{62}{\sqrt{99}}}= 2.086[/tex]
And we can use the normal standard distirbution and the complement rule to find the probability:
[tex] P(z>2.086)=1 -P(z<2.086)= 1-0.982= 0.018[/tex]
