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).

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:

[tex]X \sim N(1.65,72.26)[/tex]

Where [tex]\mu=1.65[/tex] and [tex]\sigma=72.26[/tex]

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:

[tex]P(X>a)=0.75[/tex] (a)

[tex]P(X<a)=0.25[/tex] (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:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

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

[tex]z=-0.674<\frac{a-1.65}{72.26}[/tex]

And if we solve for a we got

[tex]a=1.65 -0.674*72.26=-47.05[/tex]

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:

[tex]P(X>a)=0.25[/tex] (a)

[tex]P(X<a)=0.75[/tex] (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:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.75[/tex]

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

[tex]z=0.674<\frac{a-1.65}{72.26}[/tex]

And if we solve for a we got

[tex]a=1.65 +0.674*72.26=50.35[/tex]

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