A set of data is normally distributed with a mean of 16 and a standard deviation of 0.3. What percent of the data is between 15.2 and 16?
I had: 16-t(0.3)=15.2 and 16-t(0.3)=15.7
which came to 0.497-.341=0.156. rounded off I came up with 16%. What did I do wrong??
Thanks
I use the following z-score conversion
z = (mean - given data)/standard deviation
Your data of 15.2 translates into a z-score of -2.66666
and of course your data of 16 gives a z-score of 0
(I don't know where the 15.7 in your second calculation comes from, is it a typo?)
so you want the region between a z-score of 0 and -2.6666 which would be
.5 - .00383 = .19617
I use the following applet instead of tables, in this one you don't even have to convert to z-scores.
http://davidmlane.com/hyperstat/z_table.html
of course .5 - .00383 = .49617 and not .19617
so 49.6% of the data falls in your given range.
To find the percent of data between two values in a normal distribution, you need to calculate the area under the normal curve between those two values. In this case, you want to find the area between 15.2 and 16.
To calculate this area, you can use the Z-score formula:
Z = (X - μ) / σ
where X is the given value, μ is the mean, and σ is the standard deviation.
So first, find the Z-score for 15.2:
Z1 = (15.2 - 16) / 0.3 = -2.67
Then, find the Z-score for 16:
Z2 = (16 - 16) / 0.3 = 0
Next, you need to look up the area under the normal curve corresponding to each Z-score. You can use a Z-table or a calculator to find the cumulative probability for each Z-score.
For Z1 = -2.67, the area to the left of this Z-score is 0.0038.
For Z2 = 0, the area to the left of this Z-score is 0.5.
The percent of data between 15.2 and 16 is the difference between these cumulative probabilities:
Percent = (0.5 - 0.0038) * 100 ≈ 49.62%
So instead of 16%, the correct percentage is approximately 49.62%.
In your calculation, it appears that you used the wrong Z-scores or did not correctly obtain the cumulative probabilities from the Z-table. Make sure to use the correct Z-scores and find the appropriate cumulative probabilities to get the accurate result.