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Question:
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110.What is the probability that a domestic airfare is $550 or more (to 4 decimals)?
Answer:
0.0668
Step-by-step explanation:
We are given:
Mean cost of domestic airfare u = $385 per ticket
Standard Deviation = = $ 110
Since the distribution is normally distributed, we'll find the probability that a domestic airfare is more than or equal to $550.
Therefore
P(Airfares ≥ 550)
We are to use z score since it is normally distributed.
Let's find the equivalent z score of x at 550.
For z score, we use the expression:
Substituting figures in the expression, we have:
Therefore,
P(X ≥ 550) = P(z ≥ 1.5)
From the z table we get that P(z ≥ 1.5) = 0.0668
It can be deduced that, since P(X ≥ 550) = P(z ≥ 1.5), the probability that the domestic airfare is equal or more than $550 is 0.0668
Answer:P value = 1 - 0.9793 = 0.0207
Explanation:
we can use Z value and normal distribution to find P value. P value is the area of beyond the value of Z value
sample mean (x.bar) = $52.20
Population mean (U) = $50
Sample Standard deviation (Sd) =$ 6.10
sample (n) = 25
Z = =
Z = 2.50/1.22 = 2.049280328 = 2.049
area (normal distribution table) = 0.9793
P value = 1 - 0.9793 = 0.0207