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Each of three friends flips a coin 87 times. The results for each friend are shown in the US, FIthe relative frequency for the event "heads" for each friend. If the friends combine their results to get109 heads and 152 tails, what is the relative frequency for the event "heads"? Use pencil and paper,Suppose each friend flips a coin 870 times. Is there a value you would expect the relative frequencyfor the event "heads to be close to?Friend 1FrequencyOutcomeHeads 38Tails 49Friend 2FrequencyOutcomeHeads 43Tails 44Friend 3FrequencyoutcomeHeads 28Tails 59CThe relative frequency for the event "heads" for Friend 1 is

Each of three friends flips a coin 87 times. The results for each friend are shown-example-1
User Rami Isam
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1 Answer

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We were given that:

Each of 3 friends spin a coin 87 times

Total Trial for each friend = 87

Friend 1: Heads = 38, Tails = 49

Friend 2: Heads = 43, Tails = 44

Friend 3: Heads = 28, Tails = 59

The formula for the relative frequency of an event is given by:


Relative.Frequency=(Subgroup.Frequency)/(Total.Frequency)

We will proceed to solve for the cumulative frequency of getting "heads" as shown below:

Friend 1


\begin{gathered} Relative.Frequency=(Subgroup.Frequency)/(Total.Frequency) \\ Subgroup.Frequency=Frequency\text{ of getting''heads}^(\prime\prime)=38 \\ Total.Frequency=87 \\ Relative.Frequency=(38)/(87) \\ Relative.Frequency=0.437\approx0.44 \\ Relative.Frequency=0.44 \\ \\ \therefore Relative.Frequency=0.44 \end{gathered}

Friend 2:


\begin{gathered} Relative.Frequency=(Subgroup.Frequency)/(Total.Frequency) \\ Subgroup.Frequency=Frequency\text{ of getting''heads}^(\prime\prime)=43 \\ Total.Frequency=87 \\ Relative.Frequency=(43)/(87) \\ Relative.Frequency=0.494\approx0.49 \\ Relative.Frequency=0.49 \\ \\ \therefore Relative.Frequency=0.49 \end{gathered}

Friend 3:


\begin{gathered} Relative.Frequency=(Subgroup.Frequency)/(Total.Frequency) \\ Subgroup.Frequency=Frequency\text{ of getting''heads}^(\prime\prime)=28^{} \\ Total.Frequency=87 \\ Relative.Frequency=(28)/(87) \\ Relative.Frequency=0.322\approx0.32 \\ Relative.Frequency=0.32 \\ \\ \therefore Relative.Frequency=0.32 \end{gathered}

We will obtain the

User Khachik Sahakyan
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