Question

A given set of values is found to be a normal distribution with a mean of 140 and a standard deviation of 18.0. Find the value that is greater than 45% of the data values.

Answer 1

The value that is greater than 45% of the data values is approximately 137.84.

Explanation:

The key is transforming values from this distribution to a z-score range and finding the corresponding value using a z-score table.

We are looking for a value x which attains a critical z-score that corresponds to the (100-45)%=55-th percentile:

`z_(0.55) = (x-\mu)/(\sigma)=(x-140)/(18)\implies x = 18\cdot z_(0.55)+140`

The critical z value (from z-score table, online) is: -0.12, so:

`x = 18\cdot z_(0.55)+140=18\cdot(-0.12)+140\approx137.84`

The value that is greater than 45% of the data values is approximately 137.84.

Answer 2

Formula for Z score

`Z=(X-B)/(A)`

Where, Z is Z score for value that is greater than 45% of the data values.

X=Score=?

B=Mean =140

Standard Deviation = 18

Z score for value above 45 % of data set = 0.9987 - 0.0668=0.9319

`Z_(45 percent above)=Z_(100)-Z_(45)=0.9987-0.0668=0.9319\\\\0.9319=(X-140)/(18)\\\\ 0.9319*18=X-140\\\\X=140 +16.7742\\\\ X=156.7742`

X score for value that is greater than 45% of the data values.= 156.78 (Approx)