Exercise 1 Page P15

a We want to find the probability of landing on each color of the spinner. The experimental probability is a quotient.

P (E) = \frac{number of favorable trials}{number of trials}

For the different colors the

number of favorable trials

is given in the table. The total

number of trials

is the sum of these individual frequencies.

6 + 7 + 9 + 12 + 5 + 11 = 50

Let's find these quotients and add a column to the table as asked.

Color	Frequency	Experimental Probability
red	$6$	$\frac{6}{5 0} = 0.12$
blue	$7$	$\frac{7}{5 0} = 0.14$
yellow	$9$	$\frac{9}{5 0} = 0.18$
orange	$12$	$\frac{1 2}{5 0} = 0.24$
purple	$5$	$\frac{5}{5 0} = 0.1$
green	$11$	$\frac{1 1}{5 0} = 0.22$

b Let's now plot these values on a bar graph. The height of the bars should be proportional to the probabilities we found in part A.

c Since the spinner has sections of equal size, let's assume that landing on different colors is equally likely. Under this assumption, the theoretical probability can be calculated as a quotient.

P (E) = \frac{number of favorable outcomes}{number of possible outcomes}

In this case, the

number of possible outcomes

is the number of colors, so it is

6 .

For calculating the probability of the spinner landing on each color, the

number of favorable outcomes

1 .

Let's add these theoretical probabilities to the table.

Color	Frequency	Experimetal Probability	Theoretical Probability
red	$6$	$0.12$	$\frac{1}{6} = 0.1 \overline{6}$
blue	$7$	$0.14$	$\frac{1}{6} = 0.1 \overline{6}$
yellow	$9$	$0.18$	$\frac{1}{6} = 0.1 \overline{6}$
orange	$12$	$0.24$	$\frac{1}{6} = 0.1 \overline{6}$
purple	$5$	$0.1$	$\frac{1}{6} = 0.1 \overline{6}$
green	$11$	$0.22$	$\frac{1}{6} = 0.1 \overline{6}$

d Let's now graph the theoretical probability. The height of the bars should be proportional to the probabilities we found in part C.

e The bars on the graph of the theoretical probability are of equal height. This is the consequence of our assumption, that landing on different colors is equally likely.

The bars on the graph of the experimental probabilities are not of equal height. This indicates that the assumption of the uniform distribution may not be justified.

Subchapter links

Exercises p.15-P15