Question

App 19.study

Answer 1

Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

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:

$Pr(2) = \frac{13}{125} = 0.104$

$Pr(4) = \frac{29}{125} = 0.232$

$Pr(6) = \frac{6}{125} = 0.048$

When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

For result of roll 3 and 5

The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

$Frequency(5) = 125 * 0.296 = 37$

For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

$Frequency(1) + 103 = 125$

Collect like terms

$Frequency(1) =- 103 + 125$

$Frequency(1) =22$

The probability is then calculated as:

$Pr(1) = \frac{22}{125}$

$Pr(1) = 0.176$

So, the complete table is:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$