Question

What is the probability that a domestic airfare is $560 or more (to 4 decimals)? b. What is the probability that a domestic airfare is $260 or less (to 4 decimals)? c. What if the probability that a domestic airfare is between $320 and $500 (to 4 decimals)? d. What is the cost for the 3% highest domestic airfares? (rounded to nearest dollar)

Answer 1

Answer:

(a)

$1-P\big(\frac{X-\mu}{\sigma} < 1.809 \big) = 1 - (0.96477) = 0.03423$

(b)

$P (z < -1.047) = 0.14755$

(c)

$P(z \leq 1.238) - P(z \leq -0.47) = 0.89214 - 0.31918 = 0.5729$

Step-by-step explanation:

The first thing that you need to understand is how the random variable is distributed, it is normally distributed with mean cost $\mu = 370$  and standard deviation $\sigma = 105$.

Now for (a)

We are looking for the following probability

$P(X \geq 560)$

Here is easier to work with the complement of it, so we get

$P(X \geq 560) = 1 - P(X < 560)$

Now we need to standardize that, for standardization we use the following formula

$1-P\big(\frac{X-\mu}{\sigma} < \frac{560-\mu}{\sigma} \big) = 1-P\big(\frac{X-\mu}{\sigma} < \frac{560-370}{\105} \big) = 1-P\big(\frac{X-\mu}{\sigma} < 1.809 \big)$

And here you have to grab a standard normal table and see what is that probability.

$1-P\big(\frac{X-\mu}{\sigma} < 1.809 \big) = 1 - (0.96477) = 0.03423$

Now, if you want to interpret that result in terms of intuition, that means that the event is very unlikely.

(b)

You are looking for the following probability

$P(X \leq 260) = \\\\P\big( \frac{X-\mu}{\sigma} \leq \frac{260-\mu}{\sigma} \big)\\\\ = P\big( \frac{X-\mu}{\sigma} \leq \frac{260-370}{150} \big) \\\\= P (z < -1.047) = 0.14755$

(c)

You are looking for the following probability

$P(320 \leq X \leq 500) = P\big( \frac{320-370}{105} < z < \frac{500-370}{105} \big) \\\\= P(-0.47 < z<1.2380) = P(z \leq 1.238) - P(z \leq -0.47) = 0.89214 - 0.31918 = 0.5729$