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 0.5. A sample of 39 U.S. adults is chosen.
Is it appropriate to use the normal approximation to find the probability that more than 43% of the people in the sample have high blood pressure?
It [is or isnt?] appropriate to use the normal curve, since np=[?] [> or <] 10 and n(1-p)= [?] [> or <] 10
Is it appropriate to use the normal approximation to find the probability that more than 43% of the people in the sample have high blood pressure?
It [is or isnt?] appropriate to use the normal curve, since np=[?] [> or <] 10 and n(1-p)= [?] [> or <] 10
Answers
Answer
Find the probability that more than 43% of the people in this sample have high blood pressure.
Answer
A new sample of 79 adults is drawn. Find the probability that more than 43% of the people in this sample have high blood pressure.
Answer
Find the probability that the proportion of individuals in the sample of 79 who have high blood pressure is between .22 and .36.
Answer
Find the probability that less than 32% of the people in the sample of 79 have high blood pressure.
Answer
Would it be unusual if more than 48% of the individuals in the sample of 79 had high blood pressure?
Answered by
GPT-4o mini
To determine whether it is appropriate to use the normal approximation for the binomial distribution in this situation, we check the sample size with respect to the conditions \( np \) and \( n(1-p) \).
1. **Calculate \( p \)**:
- \( p = 0.5 \) (the proportion of U.S. adults with high blood pressure).
2. **Calculate \( n \)**:
- \( n = 39 \) (the sample size).
3. **Calculate \( np \)**:
\[
np = 39 \times 0.5 = 19.5
\]
4. **Calculate \( n(1-p) \)**:
- First, calculate \( 1 - p \):
\[
1 - p = 1 - 0.5 = 0.5
\]
- Now, calculate \( n(1-p) \):
\[
n(1 - p) = 39 \times 0.5 = 19.5
\]
Now we evaluate the conditions:
- \( np = 19.5 > 10 \)
- \( n(1 - p) = 19.5 > 10 \)
Since both conditions \( np > 10 \) and \( n(1-p) > 10 \) are satisfied, it **is appropriate** to use the normal approximation.
Final summary:
It **is appropriate** to use the normal curve, since \( np = 19.5 > 10 \) and \( n(1-p) = 19.5 > 10 \).
1. **Calculate \( p \)**:
- \( p = 0.5 \) (the proportion of U.S. adults with high blood pressure).
2. **Calculate \( n \)**:
- \( n = 39 \) (the sample size).
3. **Calculate \( np \)**:
\[
np = 39 \times 0.5 = 19.5
\]
4. **Calculate \( n(1-p) \)**:
- First, calculate \( 1 - p \):
\[
1 - p = 1 - 0.5 = 0.5
\]
- Now, calculate \( n(1-p) \):
\[
n(1 - p) = 39 \times 0.5 = 19.5
\]
Now we evaluate the conditions:
- \( np = 19.5 > 10 \)
- \( n(1 - p) = 19.5 > 10 \)
Since both conditions \( np > 10 \) and \( n(1-p) > 10 \) are satisfied, it **is appropriate** to use the normal approximation.
Final summary:
It **is appropriate** to use the normal curve, since \( np = 19.5 > 10 \) and \( n(1-p) = 19.5 > 10 \).
Answered by
GPT-4o mini
To find the probability that more than 43% of the sample of 39 U.S. adults have high blood pressure, we can use the normal approximation to the binomial distribution.
### Step 1: Define parameters
Given:
- \( n = 39 \) (sample size)
- \( p = 0.5 \) (proportion with high blood pressure)
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- Mean:
\[
\mu = p = 0.5
\]
- Standard deviation:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{39}} = \sqrt{\frac{0.25}{39}} \approx \sqrt{0.00641} \approx 0.0800
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that more than 43% of the sample has high blood pressure. First, we convert 43% to a proportion:
\[
0.43
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.5}{0.0800} = \frac{-0.07}{0.0800} \approx -0.875
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} > 0.43) \), which is equivalent to \( P(z > -0.875) \).
Using a standard normal distribution table or calculator, we find \( P(z < -0.875) \) and subtract it from 1.
From standard normal distribution tables or calculators:
\[
P(z < -0.875) \approx 0.1894
\]
Thus,
\[
P(z > -0.875) = 1 - P(z < -0.875) \approx 1 - 0.1894 = 0.8106
\]
### Conclusion
The probability that more than 43% of the people in the sample have high blood pressure is approximately **0.8106**, or **81.06%**.
### Step 1: Define parameters
Given:
- \( n = 39 \) (sample size)
- \( p = 0.5 \) (proportion with high blood pressure)
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- Mean:
\[
\mu = p = 0.5
\]
- Standard deviation:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{39}} = \sqrt{\frac{0.25}{39}} \approx \sqrt{0.00641} \approx 0.0800
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that more than 43% of the sample has high blood pressure. First, we convert 43% to a proportion:
\[
0.43
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.5}{0.0800} = \frac{-0.07}{0.0800} \approx -0.875
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} > 0.43) \), which is equivalent to \( P(z > -0.875) \).
Using a standard normal distribution table or calculator, we find \( P(z < -0.875) \) and subtract it from 1.
From standard normal distribution tables or calculators:
\[
P(z < -0.875) \approx 0.1894
\]
Thus,
\[
P(z > -0.875) = 1 - P(z < -0.875) \approx 1 - 0.1894 = 0.8106
\]
### Conclusion
The probability that more than 43% of the people in the sample have high blood pressure is approximately **0.8106**, or **81.06%**.
Answered by
GPT-4o mini
To find the probability that more than 43% of a new sample of 79 adults has high blood pressure, we can again use the normal approximation to the binomial distribution.
### Step 1: Define parameters
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that more than 43% of the sample has high blood pressure. First, convert 43% to a proportion:
\[
\hat{p} = 0.43
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.5}{0.0562} = \frac{-0.07}{0.0562} \approx -1.245
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} > 0.43) \), which is equivalent to \( P(z > -1.245) \).
Using a standard normal distribution table or calculator, we find \( P(z < -1.245) \):
\[
P(z < -1.245) \approx 0.1079
\]
Thus,
\[
P(z > -1.245) = 1 - P(z < -1.245) \approx 1 - 0.1079 = 0.8921
\]
### Conclusion
The probability that more than 43% of the people in the sample of 79 adults have high blood pressure is approximately **0.8921**, or **89.21%**.
### Step 1: Define parameters
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that more than 43% of the sample has high blood pressure. First, convert 43% to a proportion:
\[
\hat{p} = 0.43
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.5}{0.0562} = \frac{-0.07}{0.0562} \approx -1.245
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} > 0.43) \), which is equivalent to \( P(z > -1.245) \).
Using a standard normal distribution table or calculator, we find \( P(z < -1.245) \):
\[
P(z < -1.245) \approx 0.1079
\]
Thus,
\[
P(z > -1.245) = 1 - P(z < -1.245) \approx 1 - 0.1079 = 0.8921
\]
### Conclusion
The probability that more than 43% of the people in the sample of 79 adults have high blood pressure is approximately **0.8921**, or **89.21%**.
Answered by
GPT-4o mini
To find the probability that the proportion of individuals in a sample of 79 who have high blood pressure is between 0.22 and 0.36, we will again use the normal approximation to the binomial distribution.
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the proportions to z-scores
We need to find the probability that the sample proportion is between 0.22 and 0.36. We will first convert these proportions to z-scores.
#### Calculate the z-score for \( \hat{p} = 0.22 \):
\[
z_1 = \frac{0.22 - 0.5}{0.0562} = \frac{-0.28}{0.0562} \approx -4.97
\]
#### Calculate the z-score for \( \hat{p} = 0.36 \):
\[
z_2 = \frac{0.36 - 0.5}{0.0562} = \frac{-0.14}{0.0562} \approx -2.49
\]
### Step 4: Use the z-scores to find the corresponding probabilities
We need to find \( P(0.22 < \hat{p} < 0.36) \), which is equivalent to \( P(z_1 < z < z_2) \) or \( P(-4.97 < z < -2.49) \).
#### Determine probabilities from the z-table or calculator:
1. **For \( z_1 = -4.97 \)**:
- This z-score is extremely low; thus, it corresponds to a probability very close to 0, essentially \( P(z < -4.97) \approx 0.0000 \).
2. **For \( z_2 = -2.49 \)**:
- Using the z-table or calculator, we find:
\[
P(z < -2.49) \approx 0.0064
\]
### Step 5: Compute the probability for the interval
Now we can find the probability that the sample proportion is between 0.22 and 0.36:
\[
P(-4.97 < z < -2.49) = P(z < -2.49) - P(z < -4.97) \approx 0.0064 - 0.0000 = 0.0064
\]
### Conclusion
The probability that the proportion of individuals in the sample of 79 who have high blood pressure is between 0.22 and 0.36 is approximately **0.0064**, or **0.64%**.
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the proportions to z-scores
We need to find the probability that the sample proportion is between 0.22 and 0.36. We will first convert these proportions to z-scores.
#### Calculate the z-score for \( \hat{p} = 0.22 \):
\[
z_1 = \frac{0.22 - 0.5}{0.0562} = \frac{-0.28}{0.0562} \approx -4.97
\]
#### Calculate the z-score for \( \hat{p} = 0.36 \):
\[
z_2 = \frac{0.36 - 0.5}{0.0562} = \frac{-0.14}{0.0562} \approx -2.49
\]
### Step 4: Use the z-scores to find the corresponding probabilities
We need to find \( P(0.22 < \hat{p} < 0.36) \), which is equivalent to \( P(z_1 < z < z_2) \) or \( P(-4.97 < z < -2.49) \).
#### Determine probabilities from the z-table or calculator:
1. **For \( z_1 = -4.97 \)**:
- This z-score is extremely low; thus, it corresponds to a probability very close to 0, essentially \( P(z < -4.97) \approx 0.0000 \).
2. **For \( z_2 = -2.49 \)**:
- Using the z-table or calculator, we find:
\[
P(z < -2.49) \approx 0.0064
\]
### Step 5: Compute the probability for the interval
Now we can find the probability that the sample proportion is between 0.22 and 0.36:
\[
P(-4.97 < z < -2.49) = P(z < -2.49) - P(z < -4.97) \approx 0.0064 - 0.0000 = 0.0064
\]
### Conclusion
The probability that the proportion of individuals in the sample of 79 who have high blood pressure is between 0.22 and 0.36 is approximately **0.0064**, or **0.64%**.
Answered by
GPT-4o mini
To find the probability that less than 32% of the people in a sample of 79 have high blood pressure, we will again use the normal approximation to the binomial distribution.
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that less than 32% of the sample has high blood pressure. First, convert 32% to a proportion:
\[
\hat{p} = 0.32
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.32 - 0.5}{0.0562} = \frac{-0.18}{0.0562} \approx -3.20
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} < 0.32) \), which is equivalent to \( P(z < -3.20) \).
Using a standard normal distribution table or calculator, we find:
\[
P(z < -3.20) \approx 0.0007
\]
### Conclusion
The probability that less than 32% of the people in the sample of 79 have high blood pressure is approximately **0.0007**, or **0.07%**.
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Convert the sample proportion to a z-score
We want to find the probability that less than 32% of the sample has high blood pressure. First, convert 32% to a proportion:
\[
\hat{p} = 0.32
\]
Now, calculate the z-score:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.32 - 0.5}{0.0562} = \frac{-0.18}{0.0562} \approx -3.20
\]
### Step 4: Use the z-score to find the corresponding probability
We want to find \( P(\hat{p} < 0.32) \), which is equivalent to \( P(z < -3.20) \).
Using a standard normal distribution table or calculator, we find:
\[
P(z < -3.20) \approx 0.0007
\]
### Conclusion
The probability that less than 32% of the people in the sample of 79 have high blood pressure is approximately **0.0007**, or **0.07%**.
Answered by
GPT-4o mini
To determine whether it would be unusual for more than 48% of the individuals in a sample of 79 to have high blood pressure, we can use the normal approximation to the binomial distribution and calculate the corresponding z-score. We will then assess whether this event lies in a typical range (commonly defined as within 2 standard deviations from the mean).
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Calculate the z-score for \( \hat{p} = 0.48 \)
Convert 48% to a proportion and find the z-score:
\[
\hat{p} = 0.48
\]
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.48 - 0.5}{0.0562} = \frac{-0.02}{0.0562} \approx -0.355
\]
### Step 4: Assess the z-score
A z-score of around -0.355 indicates that 48% is approximately 0.355 standard deviations below the mean.
### Step 5: Find the corresponding probability
To find the probability associated with this z-score, we can look it up in the z-table or use a calculator:
\[
P(z < -0.355) \approx 0.3604
\]
This means that about 36.04% of the time, we would expect to see a sample proportion of 48% or less.
### Conclusion: Determine if it's unusual
To determine if more than 48% would be considered unusual, we consider the probability of a z-score being greater than 0.355:
\[
P(z > -0.355) = 1 - P(z < -0.355) \approx 1 - 0.3604 \approx 0.6396
\]
Approximately 63.96% of the time, we would expect to see a sample proportion greater than 48%.
Since this probability is greater than 5%, it suggests that observing more than 48% of individuals in the sample with high blood pressure is **not unusual**.
### Step 1: Define parameters
Given:
- Sample size (\( n \)) = 79
- Proportion with high blood pressure (\( p \)) = 0.5
### Step 2: Mean and standard deviation of the sampling distribution
Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution of the sample proportion (\( \hat{p} \)):
- **Mean**:
\[
\mu = p = 0.5
\]
- **Standard deviation**:
\[
\sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{79}} = \sqrt{\frac{0.25}{79}} \approx \sqrt{0.00316456} \approx 0.0562
\]
### Step 3: Calculate the z-score for \( \hat{p} = 0.48 \)
Convert 48% to a proportion and find the z-score:
\[
\hat{p} = 0.48
\]
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.48 - 0.5}{0.0562} = \frac{-0.02}{0.0562} \approx -0.355
\]
### Step 4: Assess the z-score
A z-score of around -0.355 indicates that 48% is approximately 0.355 standard deviations below the mean.
### Step 5: Find the corresponding probability
To find the probability associated with this z-score, we can look it up in the z-table or use a calculator:
\[
P(z < -0.355) \approx 0.3604
\]
This means that about 36.04% of the time, we would expect to see a sample proportion of 48% or less.
### Conclusion: Determine if it's unusual
To determine if more than 48% would be considered unusual, we consider the probability of a z-score being greater than 0.355:
\[
P(z > -0.355) = 1 - P(z < -0.355) \approx 1 - 0.3604 \approx 0.6396
\]
Approximately 63.96% of the time, we would expect to see a sample proportion greater than 48%.
Since this probability is greater than 5%, it suggests that observing more than 48% of individuals in the sample with high blood pressure is **not unusual**.
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