Find the value of x.
1 answer:
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Answer: 2.28%
Step-by-step explanation:
Given: An airport find that at the present time, the mean noise level is 103 decibels and the standard deviation is 5.4 decibels.
The distribution of noise levels(x) for all jets during takeoff over the neighborhood has a normal distribution.
The probability that jets have a noise level less than 92.2 decibels when taking off over this neighborhood will be :-
Hence, the percent of these jets have a noise level less than 92.2 decibels when taking off over this neighborhood = 2.28%
For Independent Events, P(A) × P(B) = P(A∩B)
so we have, P(A∩B) = 0.4×0.1 = 0.04
P(A') = 1 - 0.4 = 0.6
This information can be represented on a Venn diagram as shown below
P(A'∪B) means the union of everything that is not A with everything that is B
P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64
so we have, P(A∩B) = 0.4×0.1 = 0.04
P(A') = 1 - 0.4 = 0.6
This information can be represented on a Venn diagram as shown below
P(A'∪B) means the union of everything that is not A with everything that is B
P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64