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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).

User Bonzo
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Answer:

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

A very large data set (N > 10,000) has a mean value of 2.06 units and a standard-example-1
User Andy Guibert
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