Question

A very large data set (N > 10,000) has a mean value of 2.06 units and a standard deviation of 55.85 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

So in range "-35.6108 to 39.73"; 50% data set can be found

ux = 37.6704 unites

Explanation:

Step One: State the given parameters

From the question we are given that

mean Ц =2.06 unites

Standard deviation б is = 55.85 unites

Probability density → Normal

Let assume that it is symmetrically distributed about the mean

Step Two: Determine the range of the values

Our goal is to determine the range of values in which 50% of the data set should be found . Assuming that the range is from
`x_(1)` to
`x_(2)` and between these these we found 50% data set Hence from 0 up to
`x_(1)` we found 25% and up to
`x_(2)` we found 75% of the data set therefore z value corresponding to
`x_(1)` is - 0.6745 and the the z value corresponding to
`x_(2)` is +0.6745 (Refer to the z-table for normal distribution) as shown o the diagram on the first uploaded image

The formula for z is

z =
`(x-mean)/(standard deviation)`

for
`x_(1)` ,z = - 0.6745 ,substituting into the formula

`-0.6745 = (x_(1)-2.06)/(55.85)`

=> -37.67 =
`x_(1)` -2.06

=>
`x_(1)` = -35.6108 unites

for
`x_(2)` , z = 0.6745 , substituting into the formula

`0.6745 = (x_(2)-2.06)/(55.85)`

=> 37.67 =
`x_(2)` -2.06

=>
`x_(2)` = 39.73 units

So in range "-35.6108 to 39.73"; 50% data set can be found

ux = (39.73 - (-35.73) )/2

= 37.6704 unites