Respuesta :
Answer:
[tex] P(\bar X >1.7) = P(Z> \frac{1.7-1.5}{0.18}) = P(Z>1.11) = 1-P(Z<1.11) =0.1335[/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".
Solution to the problem
Let X the random variable that represent the song duration of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1.5,0.9)[/tex]
Where [tex]\mu=1.5[/tex] and [tex]\sigma=0.9[/tex]
Since the distribution for X is normal then we have that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We select a sample size of n =25 so then the deviation for the sample mean is given by:
[tex] \sigma_{\bar X}= \frac{0.9}{\sqrt{25}}= 0.18[/tex]
And we can use the z score formula given by:
[tex] z= \frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]
We want to find this probability:
[tex] P(\bar X >1.7)[/tex]
And using the z score ,the complement rule and the normal standard distribution table or excel we got:
[tex] P(\bar X >1.7) = P(Z> \frac{1.7-1.5}{0.18}) = P(Z>1.11) = 1-P(Z<1.11) =0.1335[/tex]
