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Answer:
[tex]1 \to 22 \to 0.176[/tex]
[tex]2 \to 13 \to 0.104[/tex]
[tex]3 \to 18 \to 0.144[/tex]
[tex]4 \to 29 \to 0.232[/tex]
[tex]5 \to 37 \to 0.296[/tex]
[tex]6 \to 6 \to 0.048[/tex]
Step-by-step explanation:
Given
[tex]n = 125[/tex]
See attachment for proper table
Required
Complete the table
Experimental probability is calculated as:
[tex]Pr = \frac{Frequency}{n}[/tex]
We use the above formula when the frequency is known.
For result of roll 2, 4 and 6
The frequencies are 13, 29 and 6, respectively
So, we have:
[tex]Pr(2) = \frac{13}{125} = 0.104[/tex]
[tex]Pr(4) = \frac{29}{125} = 0.232[/tex]
[tex]Pr(6) = \frac{6}{125} = 0.048[/tex]
When the frequency is to be calculated, we use:
[tex]Pr = \frac{Frequency}{n}[/tex]
[tex]Frequency = n * Pr[/tex]
For result of roll 3 and 5
The probabilities are 0.144 and 0.296, respectively
So, we have:
[tex]Frequency(3) = 125 * 0.144 = 18[/tex]
[tex]Frequency(5) = 125 * 0.296 = 37[/tex]
For roll of 1 where the frequency and the probability are not known, we use:
[tex]Total \ Frequency = 125[/tex]
So:
Frequency(1) added to others must equal 125
This gives:
[tex]Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125[/tex]
[tex]Frequency(1) + 103 = 125[/tex]
Collect like terms
[tex]Frequency(1) =- 103 + 125[/tex]
[tex]Frequency(1) =22[/tex]
The probability is then calculated as:
[tex]Pr(1) = \frac{22}{125}[/tex]
[tex]Pr(1) = 0.176[/tex]
So, the complete table is:
[tex]1 \to 22 \to 0.176[/tex]
[tex]2 \to 13 \to 0.104[/tex]
[tex]3 \to 18 \to 0.144[/tex]
[tex]4 \to 29 \to 0.232[/tex]
[tex]5 \to 37 \to 0.296[/tex]
[tex]6 \to 6 \to 0.048[/tex]
