Respuesta :
Answer:
[tex] z = \frac{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108[/tex]
And we can use the normal standard distribution table or excel and with the complement ruler we got:
[tex] P(z>-2.108) =1- P(z<-2.108) =1-0.0175= 0.9825[/tex]
Step-by-step explanation:
We know the following info given:
[tex]\mu = 997[/tex] represent the true mean
[tex]\sigma = \sqrt{3844}= 62[/tex] represent the population deviation
[tex] n = 73[/tex] represent the sample size selected
Since the sample size is large enough (n>30) we can use the central limit theorem and the sample mean would have the following distribution:
[tex]\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And for this case we want to find this probability:
[tex]P(\bar X >981.7)[/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{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108[/tex]
And we can use the normal standard distribution table or excel and with the complement ruler we got:
[tex] P(z>-2.108) =1- P(z<-2.108) =1-0.0175= 0.9825[/tex]
