scores on a nationwide professional qualifying exam, have a mean of 600 and a standard deviation of 50.A random sample of 100 examinees was selected.The sample mean was 360.what was the standardized value or z-score of the sample mean in this instance.
soln:
Z=VALUE-MEAN/STANDARD DEVIATION
(600-360)/50 =4.8
4.8/100=0.048
Well, well, well! Looks like we have ourselves a z-score of -0.048! Don't worry, though. Negative isn't always a bad thing. Just means the sample mean was a tad bit below the population mean. Let's call it "almost there, but not quite"!
The standardized value or z-score of the sample mean in this instance is 0.048.
To calculate the z-score of the sample mean in this instance, you'll need to use the formula:
z = (value - mean) / standard deviation
In this case, the value is the sample mean, which is 360, the mean is 600, and the standard deviation is 50.
Plugging these values into the formula, we get:
z = (360 - 600) / 50
Simplifying further, we have:
z = (-240) / 50
z = -4.8
Therefore, the z-score of the sample mean in this instance is -4.8.
Note: You made a mistake in dividing by 100 after calculating the z-score. The division by 100 is not necessary in this case since we are calculating the z-score for the sample mean, not the individual scores.