Introduction to a variety of useful statistical ideas and techniques - Improving the User Experience through Practical Data Analytics - page 7

After pressing the enter key, we see the probability in Figure 1.6 .
The probability is 0.9332, or, to answer the question speciically posed, 93.32%
of the users will hook everything up within 190 or less seconds.
If we wanted to ind the percentage of users who would require more than
190 seconds to hook everything up, we would simply subtract 93.32% from
100%, to get 6.68%.

1.2.2 FINDING COMPLETION TIMES OR SATISFACTION LEVELS,
OR ANYTHING ELSE, ON A NORMAL CURVE
Pretty cool so far, right? Ok, now, let's stay with the DSL installation example, but
choose another task: the time it takes to open up the installation kit and verify that
all the necessary installation components have been included in the kit. Again, we
assume that we have a normal distribution (curve) of times required to complete the
task. We know the mean equals 69 seconds and the variability factor, as measured by
the standard deviation, is 3 seconds. This indicates that there is relatively little vari-
ability, meaning that the times are somewhat similar from person to person. It also
implies that there is not a lot that the user can do to affect the task completion time,
but that the primary determination of task time is how easy it is to open the box and
ind all the parts needed.
Now, to again beef up the report and anticipate the question about how long it
takes most people to open the box and ind all the parts, you decide to determine the
range of values that 95% of the users will require (symmetric around the mean of
69 seconds) to open up the box and ind everything. In essence, you want to ind two
values of X, x ′0 and x 0, along the horizontal axis (such that the area under the curve
between the two values equals 0.95), as depicted in Figure 1.7 .

SIDEBAR: GET USED TO THE HEAVY SYMBOLS
We have labeled the two unknown values of X by the somewhat ponderous symbols of x ′ 0 and x 0.
This is based on traditional notation—and is often more onerous than need be.

X

.95

x ' x 0

µ = 69

σ = 3

FIGURE 1.7
Depiction of values sought on a normal curve, given the value of the area.