http://www.ck12.org Chapter 5. Normal Distribution
Example:
Find the probability forx≥− 0 .528.
Solution:
Right away we are at an advantage using the calculator because we do not have to round off thez−score. Enter a
normalcdfcommand from− 0 .528 to “bunches of nines”. This upper bound represents a ridiculously large upper
bound that would insure a probability of missing data being so small that it is virtually undetectable.
Remember that our answer from the table was slightly too small, so when we subtracted it from 1, it became too
large. The calculator answer of about.70125 is a more accurate approximation than the table value.
Standardizing
In most practical problems involving normal distributions, the curve will not be standardized (μ=0 andσ=1).
When using az−table, you will have to first standardize the distribution by calculating thez−score(s).
Example:
A candy company sells small bags of candy and attempts to keep the number of pieces in each bag the same, though
small differences due to random variation in the packaging process lead to different amounts in individual packages.
A quality control expert from the company has determined that the mean number of pieces in each bag is normally
distributed with a mean of 57.3 and a standard deviation of 1.2. Endy opened a bag of candy and felt he was cheated.
His bag contained only 55 candies. Does Endy have reason to complain?
Solution:
Calculate thez−score for 55.
Z=
x−μ
σ
Z=55 − 57. 3
1. 2
Z≈− 1. 911666 ...
Using Table 5.5, the probability of experiencing a value this low is approximately 0.0274. In other words, there is
about a 3% chance that you would get a bag of candy with 55 or fewer pieces, so Endy should feel cheated.
Using the graphing calculator, the results would look as follows (the ANS function has been used to avoid rounding
off thez−score):