Answer:
Event is not unusual(p>0.05).
Step-by-step explanation:
Given that :
#The sample proportion is calculated as:
#Mathematically, the z-value is the value decreased by the mean then divided the standard deviation :
#We use the normal probability table to determine the corresponding probability;
Hence, the probaility is more than 0.05, thus the event not unusual and thus this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.
Answer:
E. 0.11
Step-by-step explanation:
We have these following probabilities:
A 10% probability that a person has the flu.
A 90% probability that a person does not have the flu, just a cold.
If a person has the flu, a 99% probability of having a runny nose.
If a person just has a cold, a 90% probability of having a runny nose.
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In this problem, we have that:
What is the probability that a person has the flu, given that she has a runny nose?
P(B) is the probability that a person has the flu. So P(B) = 0.1.
P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.
P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So
What is the probability that this person has the flu?
The correct answer is:
E. 0.11