Respuesta :
Answer:
The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.
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=15,\sigma=4)[/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(15,\frac{4}{\sqrt{64}})[/tex]
Solution to the problem
We are interested on this probability
[tex]P(13.5<\bar X<16.5)[/tex]
If we apply the Z score formula to our probability we got this:
[tex]P(13.5<\bar X<16.5)=P(\frac{13.5-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{16.5-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(\frac{13.5-15}{\frac{4}{\sqrt{64}}}<Z<\frac{16.5-15}{\frac{4}{\sqrt{64}}})=P(-3<Z<3)[/tex]
And we can find this probability on this way:
[tex]P(-3<Z<3)=P(Z<3)-P(Z<-3)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.99865-0.00135=0.9973[/tex]
The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.
