Question

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 1

Answer:

a) $P(X>12)=1-\phi(2\sqrt{2})=0.24$

b)$P(x>15|x>12)=\frac{1-\phi(2\sqrt{2})}{1-\phi(\frac{\sqrt{2}}{2})}=\frac{0.002}{0.24}=0.00833$

Step-by-step explanation:

1) 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".  The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z<b)$

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:

$X \sim N(11,\sqrt{2})$  

Where $\mu=11$ and $\sigma=\sqrt{2}$

2) Part a

We are interested on this probability

$P(X>12)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>12)=P(\frac{X-\mu}{\sigma}>\frac{12-\mu}{\sigma})$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$=P(z>\frac{12-11}{\sqrt{2}})=P(z>\frac{\sqrt{2}}{2})=1-P(z<\frac{\sqrt{2}}{2})=1-P(z<0.707)$

$=1-\phi(\frac{\sqrt{2}}{2})=1-0.760=0.24$

3) Part b

For this case we want a conditional probability given by:

$P(x>15|x>12)$

And in order to solve this probability we can use the Bayes theorem given by:

$P(A|B)=\frac{P(A and B)}{P(B)}$

If we use this rule we have this:

$P(x>15|x>12)=\frac{P(x>15 and x>12)}{P(x>12)}=\frac{P(x>15)}{P(x>12)}$

The denominator is already founded on part a. So we need to find:

$P(X>15)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>15)=P(\frac{X-\mu}{\sigma}>\frac{15-\mu}{\sigma})$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$=P(z>\frac{15-11}{\sqrt{2}})=P(z>2\sqrt{2})=1-P(z<2\sqrt{2})=1-P(z<2.828)$

$=1-\phi(2\sqrt{2})=1-0.998=0.002$

$P(x>15|x>12)=\frac{1-\phi(2\sqrt{2})}{1-\phi(\frac{\sqrt{2}}{2})}=\frac{0.002}{0.24}=0.00833$

Answer 2

Using the normal distribution, it is found that:

a) 23.89% of wands are over 12 inches.

b) There is a 0.0096 = 0.96% probability it is greater than or equal to 15 inches.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is $\mu = 11$.

• The standard deviation is $\sigma = \sqrt{2} = 1.4142$.

Item a:

The proportion is 1 subtracted by the p-value of Z when X = 12, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12 - 11}{1.4142}$

$Z = 0.71$

$Z = 0.71$ has a p-value of 0.7611.

1 - 0.7611 = 0.2389.

0.2389 = 23.89% of wands are over 12 inches.

Item b:

The proportion greater than 15 inches is 1 subtracted by the p-value of Z when X = 12, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{15 - 11}{1.4142}$

$Z = 2.83$

$Z = 2.83$ has a p-value of 0.9977.

1 - 0.9977 = 0.0023.

Considering the given:

0.0023/0.2389 = 0.0096.

0.0096 = 0.96% probability it is greater than or equal to 15 inches.

More can be learned about the normal distribution at https://brainly.com/question/24663213