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

Answer:

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

ux = 37.6704 unites

Step-by-step 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 = $\frac{x-mean}{standard deviation}$

for $x_{1}$ ,z = - 0.6745 ,substituting into the formula

               $-0.6745 = \frac{x_{1}-2.06}{55.85}$

 => -37.67 =  $x_{1}$ -2.06

=> $x_{1}$ =  -35.6108 unites  

for $x_{2}$  , z = 0.6745 , substituting into the formula

            $0.6745 = \frac{x_{2}-2.06}{55.85}$

=> 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