Respuesta :
Answer:
[tex]P(78.36 \leq \bar X \leq 83.64)=P(z<2.56)-P(z<-2.56)=0.9895[/tex]
Step-by-step 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".
Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=81,\sigma=8)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(81,\frac{8}{\sqrt{60}}=1.03)[/tex]
For this case we can use the z score formula to solve the problem.The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
We want to find this probability:
[tex]P(81-2.64 \leq \bar X \leq 81+2.64)=P(78.36 \leq \bar X \leq 83.64)[/tex]
And if we replace we got:
[tex]z=\frac{83.64-81}{\frac{8}{\sqrt{60}}}=2.56[/tex]
[tex]z=\frac{78.36-81}{\frac{8}{\sqrt{60}}}=-2.56[/tex]
And we can find the probability on this way:
[tex]P(78.36 \leq \bar X \leq 83.64)=P(z<2.56)-P(z<-2.56)=0.9895[/tex]
