Given that the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42

Let X be the no of passengers who report

X is N(132, 42)

Or Z is $\frac{x-132}{42}$

a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight.

the probability that the demand will exceed the capacity of this airplane

= $P(X>183) = P(Z>1.21) =$

$=0.5-0.3869\\=0.1131$

b) the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers

= $P(80\leq x\leq 200)\\= P(-1.23\leq z\leq 1.62)\\$

=0.4474+0.3907

=0.8381

c) the probability that the demand for this flight will be less than 100 passengers

$=P(x$

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

= $P(Z>c)=0.03\\c=1.88\\X=132+1.88(42)\\=210.96$

e) 79th percentile of this distribution

= $132+0.81(42)\\= 166.02$

7 0

3 years ago

What is the value of x?

masya89 [10]

Answer:

Option D is correct.

Step-by-step explanation:

We can tell that

$4x + \frac{x}{2} + 2 + \frac{3x-4}{2} = 180$

Now, let's simplify.

=> $4x + 2 + \frac{4x-4}{2} = 180$

=> $4x + 2 + 2x-2 = 180$

=> $4x + 2x = 180$

=> $6x = 180$

=> $x = 30$

Therefore, Option D is correct.

Hoped this helped.

5 0

2 years ago

Draw a hexagon with exactly two outside diagonals

Drupady [299]

It is not possible to draw here. I am attaching the diagram along with the answer. I hope the diagram will be clear for you to understand. I also hope that this is the answer that has actually come to your desired help. You can draw it with pencil, if you want to make it perfect.

4 0

3 years ago