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Fofino [41]
3 years ago
5

Cliff wants to determine whether people prefer red apples or green apples. In order to obtain good results, he decides to conduc

t this survey at a car festival. Out of the 5000 people who attended the festival, Cliff surveys 10 attendees randomly. The survey reads, "Which do you prefer, red apples or green apples" Seven of ten people surveyed answer that they prefer red apples. From his survey, Cliff concludes that most people prefer red apples. Cliff's results may not be reliable because of _____________ problems with his study?
Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

Small sample size.

Step-by-step explanation:

The problem with Cliff's survey is that his sample size is too small. 10 out of 5,000 people represent only 0.2% of the population that attended the festival, which is too small of a sample for a reliable study.

Furthermore, since Cliff's question was to determine whether people, in general, prefer red apples or green apples, the sampling of only people who attended the festival might no be adequate.

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<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  1-P(A)+1-P(B)\\\\-P(A\cap B)\leq  1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1

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<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

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If S is a universal set then:

A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\

B can be  expressed as:

B=B\cap(A\cup A^{c})\\

If B is intersection of two disjoint sets then

P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)

Then (1) becomes

P(A\cup B) =P(A) +P(B)-P(A\cap B)\\

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\

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If B is subset of A then all elements of B lie in A so A ∩ B =B

A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})

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P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)

Therefore,

P(A)≥P(B)

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