Question

a stem-and-leaf plot for the number of touchdowns scored by all division 1a football teams is shown below. complete parts (a) through (c).

\

$$\begin{array}{r|l} 1 & 3\\ 4\\ 5\\ 7\\ 8\\ 8 \\quad \\text{key: } 1|5 = 15 \\\\ 2 & 0\\ 1\\ 2\\ 2\\ 3\\ 4\\ 5\\ 7\\ 8\\ 8\\ 9 \\\\ 3 & 0\\ 0\\ 1\\ 1\\ 1\\ 2\\ 2\\ 3\\ 3\\ 4\\ 4\\ 5\\ 5\\ 7\\ 7\\ 7\\ 8\\ 9 \\\\ 4 & 0\\ 0\\ 1\\ 2\\ 2\\ 3\\ 4\\ 5\\ 5\\ 5\\ 5\\ 6\\ 7\\ 8\\ 8\\ 9 \\\\ 5 & 0\\ 2\\ 3\\ 5\\ 5\\ 7\\ 8 \\\\ 6 & 1\\ 3\\ 6\\ 7 \\\\ 7 & \\\\ 8 & 7 \\end{array}$$

(a) if a team is selected at random, find the probability the team scored at least 32 touchdowns.
(round to three decimal places as needed.)

(b) if a team is selected at random, find the probability the team scored between 40 and 48 touchdowns inclusive.
(round to three decimal places as needed.)

(c) if a team is selected at random, find the probability the team scored more than 71 touchdowns.
(round to three decimal places as needed.)

are any of these events unusual? select all the unusual events below.

a. scoring between 40 and 48 touchdowns inclusive is unusual.
b. scoring at least 32 touchdowns is unusual.
c. scoring more than 71 touchdowns is unusual.
d. none of the events are unusual.

Explanation:

Count total observations

Using the Stem-and-Leaf Plot knowledge point, we count the total number of data values \(N\) in the plot:

• Row 1 (stem 1): 3, 4, 5, 7, 8, 8 (6 values)
• Row 2 (stem 2): 0, 1, 2, 2, 3, 4, 5, 7, 8, 8, 9 (11 values)
• Row 3 (stem 3): 0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 8, 9 (18 values)
• Row 4 (stem 4): 0, 0, 1, 2, 2, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8, 9 (16 values)
• Row 5 (stem 5): 0, 2, 3, 5, 5, 7, 8 (7 values)
• Row 6 (stem 6): 1, 3, 6, 7 (4 values)
• Row 7 (stem 7): (0 values)
• Row 8 (stem 8): 7 (1 value)

$$ N = 6 + 11 + 18 + 16 + 7 + 4 + 0 + 1 = 63 $$

Calculate probabilities

Using the Classical Probability knowledge point:

• For (a), values \(\ge 32\):
• Stem 3: 2, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 8, 9 (13 values)
• Stems 4, 5, 6, 8: \(16 + 7 + 4 + 1 = 28\) values
• Total \(\ge 32\) is \(13 + 28 = 41\)
• \(P(\ge 32) = \frac{41}{63} \approx 0.651\)
• For (b), values between 40 and 48 inclusive:
• Stem 4: 0, 0, 1, 2, 2, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8 (15 values)
• \(P(40 \le X \le 48) = \frac{15}{63} \approx 0.238\)
• For (c), values \(\ge 72\) (more than 71):
• Stem 8: 7 (1 value)
• \(P(X > 71) = \frac{1}{63} \approx 0.016\)

Identify unusual events

Using the Unusual Event Probability knowledge point:

• An event is typically considered unusual if its probability is less than or equal to 0.05.
• Comparing the calculated probabilities:
• \(P(\ge 32) = 0.651 > 0.05\) (not unusual)
• \(P(40 \le X \le 48) = 0.238 > 0.05\) (not unusual)
• \(P(X > 71) = 0.016 \le 0.05\) (unusual)
• Therefore, scoring more than 71 touchdowns is the only unusual event.

Answer:

Question 1

• A. Scoring between 40 and 48 touchdowns inclusive is unusual.
• B. Scoring at least 32 touchdowns is unusual.
• C. Scoring more than 71 touchdowns is unusual. (Correct answer)
• D. None of the events are unusual.