Question
Scores on an exam are normally distributed with a mean of 73.5 and standard deviation of 8.6. What is the probability that a random student scores 80 or higher?
Question 11 options:
0.775
0.225
0.0365
0.4325
Question 12 (1 point)
Scores on an exam are normally distributed with a mean of 73.5 and standard deviation of 8.6. What score would you need in order to be at the 90th percentile?
Question 12 options:
90.7
90
84.5
81.2
Question 13 (1 point)
Scores on an exam are normally distributed with a mean of 73.5 and standard deviation of 8.6. A sample mean will be calculated from a random sample of 16 of these exam scores. What is the probability that the sample mean will be 78 or higher?
Question 13 options:
0.018
0.300
0.389
0.982
Question 14 (1 point)
Cameron has a stack of money that consists of six $1 bills, four $5 bills, three $10 bills and nine $20 bills. If Cameron randomly selects one bill from his stack of money, what is the probability it is worth $5 or more?
Question 14 options:
4/22
16/22
12/22
10/22
Question 15 (1 point)
As gym members left their gym, they were asked how many times they had visited the gym in the last 5 days. Their responses were recorded in the given probability distribution.
x 1 2 3 4 5
P(x) 0.22 0.32 0.16 0.20 0.10
According to this data, what is the mean number of times a gym member has been to their gym in the last five days.
Question 15 options:
2.5
3
2.64
3.12
Question 11 options:
0.775
0.225
0.0365
0.4325
Question 12 (1 point)
Scores on an exam are normally distributed with a mean of 73.5 and standard deviation of 8.6. What score would you need in order to be at the 90th percentile?
Question 12 options:
90.7
90
84.5
81.2
Question 13 (1 point)
Scores on an exam are normally distributed with a mean of 73.5 and standard deviation of 8.6. A sample mean will be calculated from a random sample of 16 of these exam scores. What is the probability that the sample mean will be 78 or higher?
Question 13 options:
0.018
0.300
0.389
0.982
Question 14 (1 point)
Cameron has a stack of money that consists of six $1 bills, four $5 bills, three $10 bills and nine $20 bills. If Cameron randomly selects one bill from his stack of money, what is the probability it is worth $5 or more?
Question 14 options:
4/22
16/22
12/22
10/22
Question 15 (1 point)
As gym members left their gym, they were asked how many times they had visited the gym in the last 5 days. Their responses were recorded in the given probability distribution.
x 1 2 3 4 5
P(x) 0.22 0.32 0.16 0.20 0.10
According to this data, what is the mean number of times a gym member has been to their gym in the last five days.
Question 15 options:
2.5
3
2.64
3.12
Answers
GPT-4o mini
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**
### 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**