Answer:
The probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour is 0.0026.
Jeannie should leave her office 55 minutes early.
Step-by-step explanation:
Let <em>X</em> = time it requires Jeannie to catch her plane.
The random variable <em>X</em> is normally distributed,
.
Compute the probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour as follows:

The probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour is 0.0026.
Now, it is provided that Jeannie has 96% chance of catching her flight if she reaches under <em>x</em> minutes.
That is, P (X < x) = 0.96.
Compute the value of <em>x</em> as follows:

**Use the <em>z</em>-table to compute the value of <em>z</em>.
The value of <em>z</em> is 1.751.
Compute the value of <em>x</em> as follows:

Thus, Jeannie should leave her office 55 minutes early.