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

Answer:

Formula for Z score

$Z=\frac{X-B}{A}$

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

$Z_{45 percent above}=Z_{100}-Z_{45}=0.9987-0.0668=0.9319\\\\0.9319=\frac{X-140}{18}\\\\ 0.9319*18=X-140\\\\X=140 +16.7742\\\\ X=156.7742$

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

Answer 2

Answer:

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

Step-by-step 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:

$z_{0.55} = \frac{x-\mu}{\sigma}=\frac{x-140}{18}\implies x = 18\cdot z_{0.55}+140$

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.