Question

A recent study of 500 American males ages 18-21 found that 237 had 1 speeding ticket in the past year, 112 had 2 tickets in the past year, 17 had 3 tickets in the past year, 5 had 4 tickets in the past year, and 1 had 5 tickets in the past year. The rest of the men did not have any speeding tickets in the past year. What is the expected number of tickets an 18-21 year old American male will have in one year?

Question 2 options:

# of tickets 0 1 2 3 4 5
Frequency 128 237 112 17 5 1
Probability 0.256 0.474 0.224 0.034 0.01 0.002
# of tickets (Prob) 0.256 0.474 0.224 0.034 0.01 0.002

# of tickets 0 1 2 3 4 5
Frequency 128 237 112 17 5 1
Probability 0.256 0.474 0.224 0.034 0.01 0.002
# of tickets (Prob) 0 0.474 0.448 0.102 0.04 0.01

# of tickets 0 1 2 3 4 5
Frequency 128 237 112 17 5 1
Probability 0.256 0.474 0.224 0.034 0.01 0.002
# of tickets (Prob) 32.8 112.3 2

Answer 1

To find the expected number of tickets an 18-21 year old American male will have in one year, we can use the formula for expected value (E):

\[
E(X) = \sum (x_i \times P(x_i))
\]

Where:
- \( x_i \) is the number of tickets.
- \( P(x_i) \) is the probability of having \( x_i \) tickets.

From the data provided in your question, we have the following distributions for the number of tickets and their corresponding probabilities:

- Number of tickets (x): 0, 1, 2, 3, 4, 5
- Probability (P): 0.256, 0.474, 0.224, 0.034, 0.01, 0.002

Now we will calculate the expected value:

\[
E(X) = (0 \times 0.256) + (1 \times 0.474) + (2 \times 0.224) + (3 \times 0.034) + (4 \times 0.01) + (5 \times 0.002)
\]

Calculating each term:

- \( 0 \times 0.256 = 0 \)
- \( 1 \times 0.474 = 0.474 \)
- \( 2 \times 0.224 = 0.448 \)
- \( 3 \times 0.034 = 0.102 \)
- \( 4 \times 0.01 = 0.04 \)
- \( 5 \times 0.002 = 0.01 \)

Now add them all together:

\[
E(X) = 0 + 0.474 + 0.448 + 0.102 + 0.04 + 0.01 = 1.074
\]

Thus, the expected number of speeding tickets an 18-21 year old American male will have in one year is approximately **1.074 tickets**.