Z varies directly as x and inversely as y.

Can someone please help me with this?

aleksley [76]

Hi There!

Solution:

Double 10 = 10 * 2 = 20

Add g to the result above = 20 + g

Answer:

20 + g

Hope This Helps :)

3 0

3 years ago

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o

GREYUIT [131]

Answer:

And then we can conclude that the middle 75% values of her lap times are from a=126.835 to b=132.585

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".

2) Solution to the problem

Let X the random variable that represent the times of a population, and for this case we know the distribution for X is given by:

$X \sim N(129.71,2.5)$

Where $\mu=129.71$ and $\sigma=2.5$

Since we want the middle 75% values then on tha tails we need 1-0.75 =0.25 or 25% of the data and since the distribution is symmetric we need 12.5 % of the values on each tail.

For this part we want to find a value a and b, such that we satisfy this condition:

$P(X>b)=0.125$ (a)

$P(X$ (b)

We can use the z score formula in order to find the value a and b given by:

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

As we can see on the figure attached the z value that satisfy the condition (b) with 0.125 of the area on the left and 0.875 of the area on the right it's z=-1.15. On this case P(Z<-1.15)=0.125 and P(z>-1.15)=0.875

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-1.15$

And if we solve for a we got

$a=129.71 -1.15*2.5=126.835$

So the value of height that separates the bottom 12.5% of data from the top 87.5% is 126.835.

As we can see on the figure attached the z value that satisfy the condition (b) with 0.875 of the area on the left and 0.125 of the area on the right it's z=1.15. On this case P(Z<1.15)=0.875 and P(Z>1.15)=0.125

If we use condition (b) from previous we have this:

$P(X>b)=P(\frac{X-\mu}{\sigma}>\frac{b-\mu}{\sigma})=0.125$