Let's go through each question one by one.
Question 11
What is the probability that a random student scores 80 or higher?
To find this probability, we first need to standardize the score using the z-score formula:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \(X = 80\)
- \(\mu = 73.5\)
- \(\sigma = 8.6\)
Calculating the z-score:
\[ z = \frac{80 - 73.5}{8.6} = \frac{6.5}{8.6} \approx 0.756976 \]
Next, we can look up the z-score in the standard normal distribution table or use a calculator. The cumulative probability for \(z = 0.756976\) is approximately \(0.7764\). Therefore, the probability of scoring 80 or higher is:
\[ P(X \geq 80) = 1 - P(Z < 0.756976) \approx 1 - 0.7764 = 0.2236 \]
Thus:
Probability that a student scores 80 or higher ≈ 0.225
Question 12
What score would you need in order to be at the 90th percentile?
For the 90th percentile, we need to find the z-score that corresponds to 0.90 in the z-table, which is approximately 1.2816.
Now we convert this z-score back to a raw score using:
\[ X = \mu + z \cdot \sigma \] \[ X = 73.5 + (1.2816 \cdot 8.6) \approx 73.5 + 11.01776 \approx 84.51776 \] Rounding gives approximately 84.5.
Question 13
What is the probability that the sample mean will be 78 or higher?
Since we have a sample size \(n = 16\), the standard error (SE) is calculated as:
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{8.6}{\sqrt{16}} = \frac{8.6}{4} = 2.15 \]
Next, we calculate the z-score for the sample mean:
\[ z = \frac{\bar{X} - \mu}{SE} = \frac{78 - 73.5}{2.15} \approx \frac{4.5}{2.15} \approx 2.093 \]
Looking up this z-score, we find that the cumulative probability for \(z = 2.093\) is approximately \(0.9817\). Therefore, the probability that the sample mean will be 78 or higher is:
\[ P(\bar{X} \geq 78) = 1 - P(Z < 2.093) \approx 1 - 0.9817 \approx 0.0183 \]
Approximately 0.018.
Question 14
What is the probability it is worth $5 or more?
Cameron has:
- 6 $1 bills
- 4 $5 bills
- 3 $10 bills
- 9 $20 bills
Total number of bills:
\[ 6 + 4 + 3 + 9 = 22 \]
Number of bills worth $5 or more (4 $5 bills + 3 $10 bills + 9 $20 bills):
\[ 4 + 3 + 9 = 16 \]
Thus, the probability is:
\[ P(\text{worth } $5 \text{ or more}) = \frac{16}{22} = \frac{8}{11} \]
Question 15
What is the mean number of times a gym member has been to their gym in the last five days?
The mean \(E(X)\) is calculated using the probability distribution:
\[ E(X) = \sum (x \cdot P(x)) = (1 \cdot 0.22) + (2 \cdot 0.32) + (3 \cdot 0.16) + (4 \cdot 0.20) + (5 \cdot 0.10) \]
Calculating step by step:
\[ = 0.22 + 0.64 + 0.48 + 0.80 + 0.50 = 2.64 \]
Final Answers:
- Question 11: 0.225
- Question 12: 84.5
- Question 13: 0.018
- Question 14: 16/22
- Question 15: 2.64