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High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high...Question
High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high blood pressure is 0.3. A sample of 33 U.S. adults is chosen. Use the TI-84 Plus Calculator as needed. Round the answer to at least four decimal places.
B) A new sample of 80 adults is drawn. Find the probability that more than 38% of the people in this sample have high blood pressure.
B) A new sample of 80 adults is drawn. Find the probability that more than 38% of the people in this sample have high blood pressure.
Answers
Answer
Find the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between .24 and .35.
Answer
Find the probability that less than 26% of the people in the sample of 80 have high blood pressure.
Answer
Would it be unusual if more than42% of the individuals in the sample of 80 had high blood pressure?
Answered by
GPT-4o mini
To find the probability that more than 38% of the people in a sample of 80 adults have high blood pressure (where the population proportion \( p = 0.3 \)), we can use the normal approximation to the binomial distribution.
1. **Identify Parameters**:
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
- Sample proportion \( \hat{p} = 0.38 \)
2. **Calculate the mean and standard deviation of the sampling distribution**:
- Mean \( \mu = n \cdot p = 80 \cdot 0.3 = 24 \text{ adults} \)
- Standard deviation \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{80 \cdot 0.3 \cdot 0.7} \)
Let's compute that:
\[
\sigma = \sqrt{80 \cdot 0.3 \cdot 0.7} = \sqrt{80 \cdot 0.21} = \sqrt{16.8} \approx 4.099
\]
3. **Convert the sample proportion to a count**:
The proportion of 38% for a sample of 80 corresponds to:
\[
x = 0.38 \cdot 80 = 30.4
\]
Since we cannot have a fraction of an adult, we round up to 31.
4. **Use continuity correction**:
To find the probability that more than 38% of the sample has high blood pressure, we look for:
\[
P(X > 30)
\]
Using continuity correction, we actually want:
\[
P(X \geq 31) \approx P(X > 30.5)
\]
5. **Standardize**:
We calculate the z-score using:
\[
z = \frac{x - \mu}{\sigma} = \frac{30.5 - 24}{4.099} \approx \frac{6.5}{4.099} \approx 1.586
\]
6. **Find the probability**:
We'll use the standard normal table (or the calculator) to find:
\[
P(Z > 1.586)
\]
You can use your TI-84 calculator with the command:
```
1 - normalcdf(-1E99, 1.586)
```
This will give you the probability \( P(Z < 1.586) \). Subtract that from 1 to find \( P(Z > 1.586) \).
### Calculation on TI-84
1. Press 2nd then VARS to get to the DISTR menu.
2. Choose 2:normalcdf(.
3. Enter the values: `normalcdf(-1E99, 1.586)`.
Let’s say the normalcdf function gives us the probability of being below the calculated z-score \( P(Z < 1.586) \). For our purposes, assuming you do this calculation, you would retrieve a value.
Let’s assume the result is approximately:
\[
P(Z < 1.586) \approx 0.9418 \Rightarrow P(Z > 1.586) \approx 1 - 0.9418 = 0.0582
\]
Therefore, the final probability that more than 38% of the sample of 80 adults have high blood pressure is approximately:
\[
\boxed{0.0582}
\]
1. **Identify Parameters**:
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
- Sample proportion \( \hat{p} = 0.38 \)
2. **Calculate the mean and standard deviation of the sampling distribution**:
- Mean \( \mu = n \cdot p = 80 \cdot 0.3 = 24 \text{ adults} \)
- Standard deviation \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{80 \cdot 0.3 \cdot 0.7} \)
Let's compute that:
\[
\sigma = \sqrt{80 \cdot 0.3 \cdot 0.7} = \sqrt{80 \cdot 0.21} = \sqrt{16.8} \approx 4.099
\]
3. **Convert the sample proportion to a count**:
The proportion of 38% for a sample of 80 corresponds to:
\[
x = 0.38 \cdot 80 = 30.4
\]
Since we cannot have a fraction of an adult, we round up to 31.
4. **Use continuity correction**:
To find the probability that more than 38% of the sample has high blood pressure, we look for:
\[
P(X > 30)
\]
Using continuity correction, we actually want:
\[
P(X \geq 31) \approx P(X > 30.5)
\]
5. **Standardize**:
We calculate the z-score using:
\[
z = \frac{x - \mu}{\sigma} = \frac{30.5 - 24}{4.099} \approx \frac{6.5}{4.099} \approx 1.586
\]
6. **Find the probability**:
We'll use the standard normal table (or the calculator) to find:
\[
P(Z > 1.586)
\]
You can use your TI-84 calculator with the command:
```
1 - normalcdf(-1E99, 1.586)
```
This will give you the probability \( P(Z < 1.586) \). Subtract that from 1 to find \( P(Z > 1.586) \).
### Calculation on TI-84
1. Press 2nd then VARS to get to the DISTR menu.
2. Choose 2:normalcdf(.
3. Enter the values: `normalcdf(-1E99, 1.586)`.
Let’s say the normalcdf function gives us the probability of being below the calculated z-score \( P(Z < 1.586) \). For our purposes, assuming you do this calculation, you would retrieve a value.
Let’s assume the result is approximately:
\[
P(Z < 1.586) \approx 0.9418 \Rightarrow P(Z > 1.586) \approx 1 - 0.9418 = 0.0582
\]
Therefore, the final probability that more than 38% of the sample of 80 adults have high blood pressure is approximately:
\[
\boxed{0.0582}
\]
Answered by
GPT-4o mini
To find the probability that the proportion of individuals in a sample of 80 who have high blood pressure is between 0.24 and 0.35, we can again use the normal approximation to the binomial distribution.
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
The mean (\(\mu\)) and standard deviation (\(\sigma\)) for the sampling distribution of the sample proportion (\(\hat{p}\)) are calculated as follows:
- Mean:
\[
\mu = p = 0.3
\]
- Standard deviation:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3 \cdot (1 - 0.3)}{80}} = \sqrt{\frac{0.3 \cdot 0.7}{80}} = \sqrt{\frac{0.21}{80}} = \sqrt{0.002625} \approx 0.0512
\]
### Step 3: Convert Proportions to z-scores
We need to find the z-scores for the proportions 0.24 and 0.35.
- For \( \hat{p} = 0.24 \):
\[
z_1 = \frac{0.24 - \mu}{\sigma} = \frac{0.24 - 0.3}{0.0512} \approx \frac{-0.06}{0.0512} \approx -1.1719
\]
- For \( \hat{p} = 0.35 \):
\[
z_2 = \frac{0.35 - \mu}{\sigma} = \frac{0.35 - 0.3}{0.0512} \approx \frac{0.05}{0.0512} \approx 0.9766
\]
### Step 4: Find the Probabilities
Now we find the probabilities corresponding to these z-scores using the standard normal distribution.
Using your TI-84:
1. For \( z_1 = -1.1719 \):
- Press `2nd` then `VARS`, select `2:normalcdf(`, and enter:
```
normalcdf(-1E99, -1.1719)
```
Assuming the result is approximately \( P(Z < -1.1719) \approx 0.1206 \).
2. For \( z_2 = 0.9766 \):
- Press `2nd` then `VARS` again, and enter:
```
normalcdf(-1E99, 0.9766)
```
Assuming the result is approximately \( P(Z < 0.9766) \approx 0.8357 \).
### Step 5: Calculate the Probability that \(\hat{p}\) is between 0.24 and 0.35
The probability that the sample proportion is between 0.24 and 0.35 can be found by subtracting the two probabilities:
\[
P(0.24 < \hat{p} < 0.35) = P(Z < 0.9766) - P(Z < -1.1719) \approx 0.8357 - 0.1206 = 0.7151
\]
Thus, the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between 0.24 and 0.35 is approximately:
\[
\boxed{0.7151}
\]
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
The mean (\(\mu\)) and standard deviation (\(\sigma\)) for the sampling distribution of the sample proportion (\(\hat{p}\)) are calculated as follows:
- Mean:
\[
\mu = p = 0.3
\]
- Standard deviation:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3 \cdot (1 - 0.3)}{80}} = \sqrt{\frac{0.3 \cdot 0.7}{80}} = \sqrt{\frac{0.21}{80}} = \sqrt{0.002625} \approx 0.0512
\]
### Step 3: Convert Proportions to z-scores
We need to find the z-scores for the proportions 0.24 and 0.35.
- For \( \hat{p} = 0.24 \):
\[
z_1 = \frac{0.24 - \mu}{\sigma} = \frac{0.24 - 0.3}{0.0512} \approx \frac{-0.06}{0.0512} \approx -1.1719
\]
- For \( \hat{p} = 0.35 \):
\[
z_2 = \frac{0.35 - \mu}{\sigma} = \frac{0.35 - 0.3}{0.0512} \approx \frac{0.05}{0.0512} \approx 0.9766
\]
### Step 4: Find the Probabilities
Now we find the probabilities corresponding to these z-scores using the standard normal distribution.
Using your TI-84:
1. For \( z_1 = -1.1719 \):
- Press `2nd` then `VARS`, select `2:normalcdf(`, and enter:
```
normalcdf(-1E99, -1.1719)
```
Assuming the result is approximately \( P(Z < -1.1719) \approx 0.1206 \).
2. For \( z_2 = 0.9766 \):
- Press `2nd` then `VARS` again, and enter:
```
normalcdf(-1E99, 0.9766)
```
Assuming the result is approximately \( P(Z < 0.9766) \approx 0.8357 \).
### Step 5: Calculate the Probability that \(\hat{p}\) is between 0.24 and 0.35
The probability that the sample proportion is between 0.24 and 0.35 can be found by subtracting the two probabilities:
\[
P(0.24 < \hat{p} < 0.35) = P(Z < 0.9766) - P(Z < -1.1719) \approx 0.8357 - 0.1206 = 0.7151
\]
Thus, the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between 0.24 and 0.35 is approximately:
\[
\boxed{0.7151}
\]
Answered by
GPT-4o mini
To find the probability that less than 26% of the people in a sample of 80 have high blood pressure, we can again use the normal approximation to the binomial distribution.
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
1. **Mean (\(\mu\))**:
\[
\mu = p = 0.3
\]
2. **Standard Deviation (\(\sigma\))**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3 \cdot (1 - 0.3)}{80}} = \sqrt{\frac{0.3 \cdot 0.7}{80}} = \sqrt{\frac{0.21}{80}} = \sqrt{0.002625} \approx 0.0512
\]
### Step 3: Convert the Proportion to a z-score
For \( \hat{p} = 0.26 \):
\[
z = \frac{0.26 - \mu}{\sigma} = \frac{0.26 - 0.3}{0.0512} \approx \frac{-0.04}{0.0512} \approx -0.7812
\]
### Step 4: Find the Probability
Now we need to find \( P(Z < -0.7812) \).
Using your TI-84 calculator:
1. Press `2nd` then `VARS` to get to the DISTR menu.
2. Choose `2:normalcdf(`.
3. Enter the command as follows:
```
normalcdf(-1E99, -0.7812)
```
### Step 5: Calculation
Assuming that you perform this calculation, you would obtain:
- \( P(Z < -0.7812) \approx 0.2181 \)
Thus, the probability that less than 26% of the people in the sample of 80 have high blood pressure is approximately:
\[
\boxed{0.2181}
\]
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
1. **Mean (\(\mu\))**:
\[
\mu = p = 0.3
\]
2. **Standard Deviation (\(\sigma\))**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3 \cdot (1 - 0.3)}{80}} = \sqrt{\frac{0.3 \cdot 0.7}{80}} = \sqrt{\frac{0.21}{80}} = \sqrt{0.002625} \approx 0.0512
\]
### Step 3: Convert the Proportion to a z-score
For \( \hat{p} = 0.26 \):
\[
z = \frac{0.26 - \mu}{\sigma} = \frac{0.26 - 0.3}{0.0512} \approx \frac{-0.04}{0.0512} \approx -0.7812
\]
### Step 4: Find the Probability
Now we need to find \( P(Z < -0.7812) \).
Using your TI-84 calculator:
1. Press `2nd` then `VARS` to get to the DISTR menu.
2. Choose `2:normalcdf(`.
3. Enter the command as follows:
```
normalcdf(-1E99, -0.7812)
```
### Step 5: Calculation
Assuming that you perform this calculation, you would obtain:
- \( P(Z < -0.7812) \approx 0.2181 \)
Thus, the probability that less than 26% of the people in the sample of 80 have high blood pressure is approximately:
\[
\boxed{0.2181}
\]
Answered by
GPT-4o mini
To determine if it would be unusual for more than 42% of the individuals in a sample of 80 to have high blood pressure, we can again use the normal approximation to the binomial distribution.
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
1. **Mean (\(\mu\))**:
\[
\mu = n \cdot p = 80 \cdot 0.3 = 24
\]
2. **Standard Deviation (\(\sigma\))**:
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{80 \cdot 0.3 \cdot 0.7} = \sqrt{80 \cdot 0.21} = \sqrt{16.8} \approx 4.099
\]
### Step 3: Convert the Proportion to a Count
For \( \hat{p} = 0.42 \), we need to convert this proportion to a count of individuals in the sample:
\[
x = 0.42 \cdot 80 = 33.6
\]
Since we cannot have a fraction of an individual, we'll round this to 34.
### Step 4: Calculate the z-score
We need to find the z-score for \( x = 34 \):
\[
z = \frac{x - \mu}{\sigma} = \frac{34 - 24}{4.099} \approx \frac{10}{4.099} \approx 2.439
\]
### Step 5: Find the Probability
Now we will find \( P(Z > 2.439) \) using the standard normal distribution.
You can use your TI-84 calculator:
1. Press `2nd` then `VARS` to get to the DISTR menu.
2. Choose `2:normalcdf(`.
3. Enter the command:
```
1 - normalcdf(-1E99, 2.439)
```
### Step 6: Calculation
Assuming that you perform this calculation, let’s say we find:
- \( P(Z < 2.439) \approx 0.9925 \)
Thus:
\[
P(Z > 2.439) = 1 - 0.9925 = 0.0075
\]
### Conclusion
Since the probability \( P(Z > 2.439) \approx 0.0075 \) is very small (less than 0.01), it suggests that it would be unusual for more than 42% of the individuals in the sample of 80 to have high blood pressure.
Therefore, it would indeed be considered unusual if more than 42% of the individuals in this sample had high blood pressure.
\[
\text{Final Conclusion: } \text{Yes, it would be unusual.}
\]
### Step 1: Identify the Parameters
- Sample size \( n = 80 \)
- Population proportion \( p = 0.3 \)
### Step 2: Calculate the Mean and Standard Deviation
1. **Mean (\(\mu\))**:
\[
\mu = n \cdot p = 80 \cdot 0.3 = 24
\]
2. **Standard Deviation (\(\sigma\))**:
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{80 \cdot 0.3 \cdot 0.7} = \sqrt{80 \cdot 0.21} = \sqrt{16.8} \approx 4.099
\]
### Step 3: Convert the Proportion to a Count
For \( \hat{p} = 0.42 \), we need to convert this proportion to a count of individuals in the sample:
\[
x = 0.42 \cdot 80 = 33.6
\]
Since we cannot have a fraction of an individual, we'll round this to 34.
### Step 4: Calculate the z-score
We need to find the z-score for \( x = 34 \):
\[
z = \frac{x - \mu}{\sigma} = \frac{34 - 24}{4.099} \approx \frac{10}{4.099} \approx 2.439
\]
### Step 5: Find the Probability
Now we will find \( P(Z > 2.439) \) using the standard normal distribution.
You can use your TI-84 calculator:
1. Press `2nd` then `VARS` to get to the DISTR menu.
2. Choose `2:normalcdf(`.
3. Enter the command:
```
1 - normalcdf(-1E99, 2.439)
```
### Step 6: Calculation
Assuming that you perform this calculation, let’s say we find:
- \( P(Z < 2.439) \approx 0.9925 \)
Thus:
\[
P(Z > 2.439) = 1 - 0.9925 = 0.0075
\]
### Conclusion
Since the probability \( P(Z > 2.439) \approx 0.0075 \) is very small (less than 0.01), it suggests that it would be unusual for more than 42% of the individuals in the sample of 80 to have high blood pressure.
Therefore, it would indeed be considered unusual if more than 42% of the individuals in this sample had high blood pressure.
\[
\text{Final Conclusion: } \text{Yes, it would be unusual.}
\]
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