Suppose that the length, in inches, of a witch's or wizard's wand is a normal random variable with parameters mu = 11 and sigma^2 = 2. (a) What percentage of wands are over 12 inches? Leave you answer in terms of phi. (b) Given that a wand is greater than 12 inches, what is the probability it is greater than or equal to 15 inches? Leave you answer in terms of phi.

Answer:a) [tex]P(X>12)=1-\phi(2\sqrt{2})=0.24[/tex]b)[tex]P(x>15|x>12)=\frac{1-\phi(2\sqrt{2})}{1-\phi(\frac{\sqrt{2}}{2})}=\frac{0.002}{0.24}=0.00833[/tex]Step-by-step explanation:1) Previous conceptsNormal 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".  The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]Let X the random variable that represent the length of a population, and for this case we know the distribution for X is given by:[tex]X \sim N(11,\sqrt{2})[/tex]  Where [tex]\mu=11[/tex] and [tex]\sigma=\sqrt{2}[/tex]2) Part aWe are interested on this probability[tex]P(X>12)[/tex]And the best way to solve this problem is using the normal standard distribution and the z score given by:[tex]z=\frac{x-\mu}{\sigma}[/tex]If we apply this formula to our probability we got this:[tex]P(X>12)=P(\frac{X-\mu}{\sigma}>\frac{12-\mu}{\sigma})[/tex]And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  [tex]=P(z>\frac{12-11}{\sqrt{2}})=P(z>\frac{\sqrt{2}}{2})=1-P(z<\frac{\sqrt{2}}{2})=1-P(z<0.707)[/tex][tex]=1-\phi(\frac{\sqrt{2}}{2})=1-0.760=0.24[/tex]3) Part bFor this case we want a conditional probability given by:[tex]P(x>15|x>12)[/tex]And in order to solve this probability we can use the Bayes theorem given by:[tex]P(A|B)=\frac{P(A and B)}{P(B)}[/tex]If we use this rule we have this:[tex]P(x>15|x>12)=\frac{P(x>15 and x>12)}{P(x>12)}=\frac{P(x>15)}{P(x>12)}[/tex]The denominator is already founded on part a. So we need to find:[tex]P(X>15)[/tex]And the best way to solve this problem is using the normal standard distribution and the z score given by:[tex]z=\frac{x-\mu}{\sigma}[/tex]If we apply this formula to our probability we got this:[tex]P(X>15)=P(\frac{X-\mu}{\sigma}>\frac{15-\mu}{\sigma})[/tex]And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  [tex]=P(z>\frac{15-11}{\sqrt{2}})=P(z>2\sqrt{2})=1-P(z<2\sqrt{2})=1-P(z<2.828)[/tex][tex]=1-\phi(2\sqrt{2})=1-0.998=0.002[/tex][tex]P(x>15|x>12)=\frac{1-\phi(2\sqrt{2})}{1-\phi(\frac{\sqrt{2}}{2})}=\frac{0.002}{0.24}=0.00833[/tex]