Answer :
Answer:
[tex] z = \frac{135.7-135}{\frac{10}{\sqrt{43}}}= 0.459[/tex]
And we can find this probability with this way:
[tex] P(z> 0.459) =1 -P(z<0.459)= 1-0.677= 0.323[/tex]
Step-by-step explanation:
For this case w ehave the following info given:
[tex] \mu = 135[/tex] represent the sample mean
[tex] \sigma^2 = 100[/tex] represent the sample variance
[tex] \sigma =\sqrt{100}=10[/tex] represent the deviation
[tex] n = 43[/tex] represent the sample mean selected
For this case we want to find the following probability:
[tex] P(\bar X >135.7)[/tex]
And since the sample size is large enough we can use the following distribution for the sample mean:
[tex]\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}}) [/tex]
And we can use the z score formula given by:
[tex] z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And we can find the z score for the value 135.7 and we got:
[tex] z = \frac{135.7-135}{\frac{10}{\sqrt{43}}}= 0.459[/tex]
And we can find this probability with this way:
[tex] P(z> 0.459) =1 -P(z<0.459)= 1-0.677= 0.323[/tex]