Respuesta :
Answer:
[tex] P( \bar X >104) = P(Z > \frac{104-100}{\frac{15}{\sqrt{50}}}) = P(Z>1.886)[/tex]
And we can use the complement rule and the normal standard distribution or excel and we got:
[tex] P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03[/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 IQ of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We select a sample of n = 50 and we want to find the probability that:
[tex]P(\bar X >104)[/tex]
Since the distribution for X is normal then we know 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]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} [/tex]
And using this formula we got:
[tex] P( \bar X >104) = P(Z > \frac{104-100}{\frac{15}{\sqrt{50}}}) = P(Z>1.886)[/tex]
And we can use the complement rule and the normal standard distribution or excel and we got:
[tex] P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03[/tex]
