Obstacle Course
Suppose a group of students are running through a timed obstacle course for class. Their recorded times for course completion form a normal distribution with a mean of 8.0 minutes and a standard deviation of 2.5 minutes. What is the probability that a student (selected at random) takes longer than 15.5 minutes to finish the course? Refer to the shown Unit Normal Distribution table.
Expand Hint
For any normal distribution, the table for the unit normal distribution can be utilized via the following transformation:
$$$z=\frac{x-\mu }{\sigma }$$$
where
$$z$$
is the “Z” score associated with the unit normal distribution table,
$$\mu$$
is the mean,
$$\sigma$$
is the standard deviation, and
$$x$$
is the desired region under the bell curve.
Hint 2
For any normal distribution, the table for the unit normal distribution can be utilized via the following transformation:
$$$z=\frac{x-\mu }{\sigma }$$$
where
$$z$$
is the “Z” score associated with the unit normal distribution table,
$$\mu$$
is the mean,
$$\sigma$$
is the standard deviation, and
$$x$$
is the desired region under the bell curve.
$$$z=\frac{x-\mu }{\sigma }=\frac{15.5-8}{2.5}=3.00$$$
Referring to the Unit Normal Distribution Table:
Because the label for rows contains the integer part and the first decimal place of Z, and the label for columns contains the second decimal place of Z, “.99865” is the corresponding distribution for
$$Z=3.00$$
. This represents the probability of selecting a student whose course completion time is under 15.5 minutes. To find the probability of selecting a student who takes
LONGER
than 15.5 minutes:
$$$1-.99865=0.00135\:probability$$$
0.00135 probability
Time Analysis
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Similar Problems from FE Sub Section: Normal Distribution (Gaussian Distribution)
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621. Z Score
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