A spinner for a game has the numbers 1 through 7 on it. The spinner is equally likely to stop on each of the 7 numbers. The numbers 1 and 2 are colored red. The numbers 3, 4, 5, 6 and 7 are colored black.

Question

The spinner is spun one time. What is the probability the spinner stops on a number that is even or red?

Question 7 options:

2/7

3/7

4/7

5/7

Question 8 (1 point)

A standard six-sided die has sides numbered 1, 2, 3, 4, 5, and 6. This die will be rolled eight times. On each roll of the die, either the number one is rolled, or the number one is not rolled. Find the probability the number one is rolled on exactly two of the eight rolls?

Question 8 options:

0.028

0.260

0.333

0.25

Question 9 (1 point)

The probability that Arianna makes a free throw on any one attempt is 0.63. If she attempts 20 free throws, and each attempt is independent, how many of the 20 free throws do we expect her to make?

Question 9 options:

10

6.3

12.6

14

Question 10 (1 point)

At one-year of age, the lengths of female American alligators are normally distributed with a population mean length of 49cm and a standard deviation of 4 cm. Between what two lengths will approximately 95% of all one-year old female American alligators be?

Question 10 options:

45 and 53 cm

41 and 57 cm

37 and 61 cm

49 and 61 cm

Answer 1

Let's go through each question one by one.

Question 7: Probability the spinner stops on a number that is even or red

The numbers on the spinner are: 1 (red), 2 (red), 3 (black), 4 (black), 5 (black), 6 (black), 7 (black).

Even numbers: 2, 4, 6

• Total even numbers: 3

Red numbers: 1, 2

• Total red numbers: 2

Now, let's find the union of the two events (even numbers or red numbers). The even numbers that are counted amongst the red ones is 2.

Using the formula for the probability of the union (P(A or B) = P(A) + P(B) - P(A and B)):

• P(even) = 3/7
• P(red) = 2/7
• P(even and red) = 1/7 (since the only overlap is the number 2)

So: \[ P(\text{even or red}) = P(\text{even}) + P(\text{red}) - P(\text{even and red}) \] \[ = \frac{3}{7} + \frac{2}{7} - \frac{1}{7} = \frac{4}{7} \]

Answer for Question 7: 4/7

---

Question 8: Probability of rolling exactly two 1's in eight rolls of a die

Here, we can use the binomial probability formula:

\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:

• \( n = 8 \) (the number of trials)
• \( k = 2 \) (the number of successes)
• \( p = \frac{1}{6} \) (the probability of rolling a 1)
• \( (1-p) = \frac{5}{6} \) (the probability of not rolling a 1)

Calculating:

- Calculate \(\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28\)

Substituting into the formula:

\[ P(X=2) = 28 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{6} \]

Calculating:

• \(\left(\frac{1}{6}\right)^2 = \frac{1}{36}\)
• \(\left(\frac{5}{6}\right)^{6} = \left(\frac{5^6}{6^6}\right) = \frac{15625}{46656}\)

Now multiplying:

\[ P(X=2) = 28 \times \frac{1}{36} \times \frac{15625}{46656} \]

After calculating:

\[ P(X=2) \approx 0.260 \]

Answer for Question 8: 0.260

---

Question 9: Expected number of free throws Arianna makes

The expected number of successes in a binomial distribution can be calculated using:

\[ E(X) = n \cdot p \]

where:

• \( n = 20 \) (the number of trials)
• \( p = 0.63 \) (the probability of success)

Calculating:

\[ E(X) = 20 \cdot 0.63 = 12.6 \]

Answer for Question 9: 12.6

---

Question 10: Lengths of female American alligators in cm

For a normal distribution, approximately 95% of data lies within 2 standard deviations from the mean.

Given:

• Mean \( \mu = 49 , \text{cm} \)
• Standard deviation \( \sigma = 4 , \text{cm} \)

The range for 95% would be calculated as follows:

Lower Bound: \[ \mu - 2\sigma = 49 - 2(4) = 49 - 8 = 41 , \text{cm} \]

Upper Bound: \[ \mu + 2\sigma = 49 + 2(4) = 49 + 8 = 57 , \text{cm} \]

Thus, approximately 95% of one-year-old female American alligators will be between 41 cm and 57 cm.

Answer for Question 10: 41 and 57 cm