Question

A very large data set (N > 10,000) has a mean value of 1.65 units and a standard deviation of 72.26 units. Determine the range of values in which 50% of the data set should be found assuming normal probability density (find ux in the equation xi = x' ± ux).

Answer 1

Answer:

$z=-0.674$

And if we solve for a we got

$a=1.65 -0.674*72.26=-47.05$

$z=0.674$

And if we solve for a we got

$a=1.65 +0.674*72.26=50.35$

So then the limits where 50% of the data lies are -47.05 and 50.35

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 variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(1.65,72.26)$  

Where $\mu=1.65$ and $\sigma=72.26$

For this case we want the limits for the 50% of the values.

So on the tails of the distribution we need the other 50% of the data, and on ach tail we need to have 25% since the distribution is symmetric.

Lower tail

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

$P(X>a)=0.75$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

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:

$z=-0.674$

And if we solve for a we got

$a=1.65 -0.674*72.26=-47.05$

So the value of height that separates the bottom 25% of data from the top 75% is -47.05.  

Upper tail

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

$P(X>a)=0.25$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

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:

$z=0.674$

And if we solve for a we got

$a=1.65 +0.674*72.26=50.35$

So the value of height that separates the bottom 75% of data from the top 25% is 50.35.  

So then the limits where 50% of the data lies are -47.05 and 50.35