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TonyHaas07
25.10.2019 •
Mathematics
You calculated probabilities involving various blood types. some of your answers depended on the assumption that the outcomes described were disjoint; that is, they could not both happen at the same time. other answers depended on the assumption that the events were independent; that is, the occurrence of one of them doesn't affect the probability of the other. do you understand the difference between disjoint and independent? a) if you examine one person, are the events that the person is type a and that the same person is type b disjoint , independent, or neither? b) if you examine two people , are the events that the first is type a and the second type b disjoint , independent, or neither? c) can disjoint events ever be independent? explain
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Ответ:
a) The events are disjoint, because a person cannot have blood type A and blood type B at the same time.
b) The events are independent, because the probability of having blood type does not depend on other persod having blood type B.
c) Yes, disjoint events are independent if and only if the probability of some of the events is zero.
Step-by-step explanation:
a) Let the events be:
A: "the person has blood type A
B: the person has blood type B
As a person can only have one blood type these events are disjoint.
b) If we examine two people, let the events be that the first is blood Type A and the second blood Type B:
The events are independent because the probability of the first person having blood type A does not influence the probability of the second person having blood type B (unless the two people are related).
c) Two events are independent if and only if:
P(A∩B)=P(A)×P(B)
Given the events A and B which are disjoint:
P(A∩B)=0
Hence, A and B could be independent events if and only if the probability of one of the events is 0.
Ответ:
The probability that the mean receipt for dinner is between $24.50 and $25.50 is
P(24.50≤ x ≤ 25.50)= 0.8324
Step-by-step explanation:
Step(i):-
Given mean of Population(μ) = $25
Given standard deviation of the Population (σ) = 2.30
Sample size 'n' = 40
Let 'X' be the random variable in normal distribution
Given X₁ = 24.50
Given X₂ = 25.50
Step(ii):-
The probability that the mean receipt for dinner is between $24.50 and $25.50
P(x₁≤ x ≤ x₂) = P(Z₁≤ z ≤ Z₂) = A(Z₂) + A(z₁)
P(24.50≤ x ≤ 25.50) = P(-1.375≤ z ≤ 1.375) = A(1.375) + A(1.375)
P(24.50≤ x ≤ 25.50)= 2 A( 1.375)
= 2 × 0.4162 ( from normal table)
= 0.8324
Final answer:-
The probability that the mean receipt for dinner is between $24.50 and $25.50 is 0.8324