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 0.4. A sample of 40 U.S. adults is chosen.
Is it appropriate to use the normal approximation to find the probability that more than 42% of the people in the sample have high blood pressure?
It [is or isnt?] appropriate 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 42% of the people in the sample have high blood pressure?
It [is or isnt?] appropriate since np = [?] [> or <] 10, and n(1-p)= [?] [> or <] 10
Answers
GPT-4o mini
To determine whether it is appropriate to use the normal approximation for the binomial distribution when sampling U.S. adults for high blood pressure, we need to check the conditions \( np \) and \( n(1-p) \).
Given:
- \( n = 40 \) (sample size)
- \( p = 0.4 \) (proportion of U.S. adults with high blood pressure)
Now, we can calculate:
1. \( np \):
\[
np = 40 \times 0.4 = 16
\]
2. \( n(1-p) \):
\[
n(1-p) = 40 \times (1 - 0.4) = 40 \times 0.6 = 24
\]
Now we can evaluate:
- \( np = 16 \) \( [>\] 10 \)
- \( n(1-p) = 24 \) \( [>\] 10 \)
Since both \( np \) and \( n(1-p) \) are greater than 10, it **is appropriate** to use the normal approximation.
Final answer:
It **is** appropriate since \( np = 16 \) \( [>] 10 \), and \( n(1-p)= 24 \) \( [>] 10 \).
Given:
- \( n = 40 \) (sample size)
- \( p = 0.4 \) (proportion of U.S. adults with high blood pressure)
Now, we can calculate:
1. \( np \):
\[
np = 40 \times 0.4 = 16
\]
2. \( n(1-p) \):
\[
n(1-p) = 40 \times (1 - 0.4) = 40 \times 0.6 = 24
\]
Now we can evaluate:
- \( np = 16 \) \( [>\] 10 \)
- \( n(1-p) = 24 \) \( [>\] 10 \)
Since both \( np \) and \( n(1-p) \) are greater than 10, it **is appropriate** to use the normal approximation.
Final answer:
It **is** appropriate since \( np = 16 \) \( [>] 10 \), and \( n(1-p)= 24 \) \( [>] 10 \).
Find the probability that more than 42% of the people in this sample have high blood pressure.
GPT-4o mini
To find the probability that more than 42% of the sample of 40 U.S. adults have high blood pressure, we will use the normal approximation to the binomial distribution.
### Step-by-step calculation:
1. **Determine the mean (\( \mu \)) and standard deviation (\( \sigma \))** of the binomial distribution.
- The mean \( \mu \) of a binomial distribution is given by:
\[
\mu = n \cdot p = 40 \cdot 0.4 = 16
\]
- The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{40 \cdot 0.4 \cdot 0.6} = \sqrt{9.6} \approx 3.1
\]
2. **Convert the proportion to the number of individuals**.
- 42% of 40 adults is:
\[
0.42 \cdot 40 = 16.8
\]
- We need to find \( P(X > 16.8) \). Since we're dealing with a discrete distribution, we can use the continuity correction factor, essentially considering \( P(X > 16.8) \) as \( P(X \geq 17) \).
3. **Convert to the standard normal variable (Z-score)**.
- The Z-score is given by:
\[
Z = \frac{X - \mu}{\sigma}
\]
- Here, we want \( X = 17 \):
\[
Z = \frac{17 - 16}{3.1} \approx \frac{1}{3.1} \approx 0.3226
\]
4. **Use the Z-score to find the probability**.
- We want \( P(X > 17) = 1 - P(Z \leq 0.3226) \).
- Looking up \( Z = 0.3226 \) in the standard normal distribution table (or using a calculator), we find approximately:
\[
P(Z \leq 0.3226) \approx 0.6260
\]
- Thus, the probability that more than 42% of the sample have high blood pressure is:
\[
P(X > 17) = 1 - P(Z \leq 0.3226) \approx 1 - 0.6260 \approx 0.3740
\]
### Final Answer:
The probability that more than 42% of the people in this sample have high blood pressure is approximately **0.3740**, or **37.40%**.
### Step-by-step calculation:
1. **Determine the mean (\( \mu \)) and standard deviation (\( \sigma \))** of the binomial distribution.
- The mean \( \mu \) of a binomial distribution is given by:
\[
\mu = n \cdot p = 40 \cdot 0.4 = 16
\]
- The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{40 \cdot 0.4 \cdot 0.6} = \sqrt{9.6} \approx 3.1
\]
2. **Convert the proportion to the number of individuals**.
- 42% of 40 adults is:
\[
0.42 \cdot 40 = 16.8
\]
- We need to find \( P(X > 16.8) \). Since we're dealing with a discrete distribution, we can use the continuity correction factor, essentially considering \( P(X > 16.8) \) as \( P(X \geq 17) \).
3. **Convert to the standard normal variable (Z-score)**.
- The Z-score is given by:
\[
Z = \frac{X - \mu}{\sigma}
\]
- Here, we want \( X = 17 \):
\[
Z = \frac{17 - 16}{3.1} \approx \frac{1}{3.1} \approx 0.3226
\]
4. **Use the Z-score to find the probability**.
- We want \( P(X > 17) = 1 - P(Z \leq 0.3226) \).
- Looking up \( Z = 0.3226 \) in the standard normal distribution table (or using a calculator), we find approximately:
\[
P(Z \leq 0.3226) \approx 0.6260
\]
- Thus, the probability that more than 42% of the sample have high blood pressure is:
\[
P(X > 17) = 1 - P(Z \leq 0.3226) \approx 1 - 0.6260 \approx 0.3740
\]
### Final Answer:
The probability that more than 42% of the people in this sample have high blood pressure is approximately **0.3740**, or **37.40%**.
A new sample of 78 adults is drawn. Find the probability that more than 41% of the people in this sample have high blood pressure.
GPT-4o mini
To find the probability that more than 41% of a new sample of 78 adults have high blood pressure, we will again utilize the normal approximation to the binomial distribution.
### Step-by-step calculation:
1. **Determine the mean (\( \mu \)) and standard deviation (\( \sigma \))** of the binomial distribution.
- Given:
- Sample size \( n = 78 \)
- Population proportion \( p = 0.4 \)
- The mean \( \mu \) is calculated as follows:
\[
\mu = n \cdot p = 78 \cdot 0.4 = 31.2
\]
- The standard deviation \( \sigma \) is calculated as:
\[
\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{78 \cdot 0.4 \cdot 0.6} = \sqrt{18.72} \approx 4.33
\]
2. **Convert the proportion to the number of individuals**.
- 41% of 78 adults is:
\[
0.41 \cdot 78 = 31.98
\]
- We need to find \( P(X > 31.98) \). Using a continuity correction, we consider this as \( P(X \geq 32) \).
3. **Convert to the standard normal variable (Z-score)**.
- The Z-score is given by:
\[
Z = \frac{X - \mu}{\sigma}
\]
- Here, we want \( X = 32 \):
\[
Z = \frac{32 - 31.2}{4.33} \approx \frac{0.8}{4.33} \approx 0.184
\]
4. **Use the Z-score to find the probability**.
- We want \( P(X > 32) = 1 - P(Z \leq 0.184) \).
- Looking up \( Z = 0.184 \) in the standard normal distribution table (or using a calculator), we find approximately:
\[
P(Z \leq 0.184) \approx 0.572
\]
- Thus, the probability that more than 41% of the sample have high blood pressure is:
\[
P(X > 32) = 1 - P(Z \leq 0.184) \approx 1 - 0.572 \approx 0.428
\]
### Final Answer:
The probability that more than 41% of the people in this new sample of 78 adults have high blood pressure is approximately **0.428**, or **42.8%**.
### Step-by-step calculation:
1. **Determine the mean (\( \mu \)) and standard deviation (\( \sigma \))** of the binomial distribution.
- Given:
- Sample size \( n = 78 \)
- Population proportion \( p = 0.4 \)
- The mean \( \mu \) is calculated as follows:
\[
\mu = n \cdot p = 78 \cdot 0.4 = 31.2
\]
- The standard deviation \( \sigma \) is calculated as:
\[
\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{78 \cdot 0.4 \cdot 0.6} = \sqrt{18.72} \approx 4.33
\]
2. **Convert the proportion to the number of individuals**.
- 41% of 78 adults is:
\[
0.41 \cdot 78 = 31.98
\]
- We need to find \( P(X > 31.98) \). Using a continuity correction, we consider this as \( P(X \geq 32) \).
3. **Convert to the standard normal variable (Z-score)**.
- The Z-score is given by:
\[
Z = \frac{X - \mu}{\sigma}
\]
- Here, we want \( X = 32 \):
\[
Z = \frac{32 - 31.2}{4.33} \approx \frac{0.8}{4.33} \approx 0.184
\]
4. **Use the Z-score to find the probability**.
- We want \( P(X > 32) = 1 - P(Z \leq 0.184) \).
- Looking up \( Z = 0.184 \) in the standard normal distribution table (or using a calculator), we find approximately:
\[
P(Z \leq 0.184) \approx 0.572
\]
- Thus, the probability that more than 41% of the sample have high blood pressure is:
\[
P(X > 32) = 1 - P(Z \leq 0.184) \approx 1 - 0.572 \approx 0.428
\]
### Final Answer:
The probability that more than 41% of the people in this new sample of 78 adults have high blood pressure is approximately **0.428**, or **42.8%**.
Find the probability that the proportion of individuals in the sample of 78 who have high blood pressure is between .21 and .32.
GPT-4o mini
To find the probability that the proportion of individuals in a sample of 78 who have high blood pressure is between 0.21 and 0.32, we can use the normal approximation to the binomial distribution. Here's how to calculate it step by step:
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution
1. **Population proportion**: Let \( p = 0.4 \).
2. **Sample size**: \( n = 78 \).
- The mean of the sampling distribution of the sample proportion (\( \mu \)) is given by:
\[
\mu = p = 0.4
\]
- The standard deviation of the sampling distribution (\( \sigma \)) is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert proportions to Z-scores
Next, convert the proportions 0.21 and 0.32 into Z-scores.
- For \( p_1 = 0.21 \):
\[
Z_1 = \frac{p_1 - \mu}{\sigma} = \frac{0.21 - 0.4}{0.0555} \approx \frac{-0.19}{0.0555} \approx -3.42
\]
- For \( p_2 = 0.32 \):
\[
Z_2 = \frac{p_2 - \mu}{\sigma} = \frac{0.32 - 0.4}{0.0555} \approx \frac{-0.08}{0.0555} \approx -1.44
\]
### Step 3: Find probabilities using Z-scores
Now, use the Z-scores to find the corresponding probabilities from the standard normal distribution:
- Looking up \( Z_1 \approx -3.42 \), we find:
\[
P(Z \leq -3.42) \approx 0.0003 \quad (\text{very small probability})
\]
- Looking up \( Z_2 \approx -1.44 \), we find:
\[
P(Z \leq -1.44) \approx 0.0753
\]
### Step 4: Calculate the probability that the proportion is between 0.21 and 0.32
Finally, we find the probability that the proportion is between 0.21 and 0.32:
\[
P(0.21 < p < 0.32) = P(Z \leq -1.44) - P(Z \leq -3.42) \approx 0.0753 - 0.0003 = 0.0750
\]
### Final Answer
The probability that the proportion of individuals in the sample of 78 who have high blood pressure is between 0.21 and 0.32 is approximately **0.0750**, or **7.50%**.
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the sampling distribution
1. **Population proportion**: Let \( p = 0.4 \).
2. **Sample size**: \( n = 78 \).
- The mean of the sampling distribution of the sample proportion (\( \mu \)) is given by:
\[
\mu = p = 0.4
\]
- The standard deviation of the sampling distribution (\( \sigma \)) is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert proportions to Z-scores
Next, convert the proportions 0.21 and 0.32 into Z-scores.
- For \( p_1 = 0.21 \):
\[
Z_1 = \frac{p_1 - \mu}{\sigma} = \frac{0.21 - 0.4}{0.0555} \approx \frac{-0.19}{0.0555} \approx -3.42
\]
- For \( p_2 = 0.32 \):
\[
Z_2 = \frac{p_2 - \mu}{\sigma} = \frac{0.32 - 0.4}{0.0555} \approx \frac{-0.08}{0.0555} \approx -1.44
\]
### Step 3: Find probabilities using Z-scores
Now, use the Z-scores to find the corresponding probabilities from the standard normal distribution:
- Looking up \( Z_1 \approx -3.42 \), we find:
\[
P(Z \leq -3.42) \approx 0.0003 \quad (\text{very small probability})
\]
- Looking up \( Z_2 \approx -1.44 \), we find:
\[
P(Z \leq -1.44) \approx 0.0753
\]
### Step 4: Calculate the probability that the proportion is between 0.21 and 0.32
Finally, we find the probability that the proportion is between 0.21 and 0.32:
\[
P(0.21 < p < 0.32) = P(Z \leq -1.44) - P(Z \leq -3.42) \approx 0.0753 - 0.0003 = 0.0750
\]
### Final Answer
The probability that the proportion of individuals in the sample of 78 who have high blood pressure is between 0.21 and 0.32 is approximately **0.0750**, or **7.50%**.
Find the probability that less than 22% of the people in the sample of 78 have high blood pressure.
GPT-4o mini
To find the probability that less than 22% of the people in a sample of 78 adults have high blood pressure, we will again use the normal approximation to the binomial distribution.
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \))
1. **Population proportion**: The population proportion is \( p = 0.4 \).
2. **Sample size**: The sample size is \( n = 78 \).
- The mean \( \mu \) of the sampling distribution of the sample proportion is:
\[
\mu = p = 0.4
\]
- The standard deviation \( \sigma \) of the sampling distribution is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert the proportion to a Z-score
Next, convert the proportion of 22% into a Z-score.
- For \( p = 0.22 \):
\[
Z = \frac{p - \mu}{\sigma} = \frac{0.22 - 0.4}{0.0555} \approx \frac{-0.18}{0.0555} \approx -3.24
\]
### Step 3: Find the probability using the Z-score
Now, we will find the probability corresponding to that Z-score.
- Using a standard normal distribution table or a calculator for \( Z = -3.24 \):
\[
P(Z \leq -3.24) \approx 0.0006
\]
### Final Answer
The probability that less than 22% of the people in the sample of 78 have high blood pressure is approximately **0.0006**, or **0.06%**.
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \))
1. **Population proportion**: The population proportion is \( p = 0.4 \).
2. **Sample size**: The sample size is \( n = 78 \).
- The mean \( \mu \) of the sampling distribution of the sample proportion is:
\[
\mu = p = 0.4
\]
- The standard deviation \( \sigma \) of the sampling distribution is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert the proportion to a Z-score
Next, convert the proportion of 22% into a Z-score.
- For \( p = 0.22 \):
\[
Z = \frac{p - \mu}{\sigma} = \frac{0.22 - 0.4}{0.0555} \approx \frac{-0.18}{0.0555} \approx -3.24
\]
### Step 3: Find the probability using the Z-score
Now, we will find the probability corresponding to that Z-score.
- Using a standard normal distribution table or a calculator for \( Z = -3.24 \):
\[
P(Z \leq -3.24) \approx 0.0006
\]
### Final Answer
The probability that less than 22% of the people in the sample of 78 have high blood pressure is approximately **0.0006**, or **0.06%**.
Would it be unusual if more than 32% of the individuals in the sample of 78 had high blood pressure?
it [would or wouldnt] since the probability is [?]
it [would or wouldnt] since the probability is [?]
GPT-4o mini
To determine whether it would be unusual for more than 32% of the individuals in a sample of 78 to have high blood pressure, we will proceed with the following analysis:
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \))
1. **Population proportion**: The population proportion is \( p = 0.4 \).
2. **Sample size**: The sample size is \( n = 78 \).
- The mean \( \mu \) of the sampling distribution of the sample proportion is:
\[
\mu = p = 0.4
\]
- The standard deviation \( \sigma \) of the sampling distribution is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert the proportion to a Z-score
Next, convert the proportion of 32% into a Z-score.
- For \( p = 0.32 \):
\[
Z = \frac{p - \mu}{\sigma} = \frac{0.32 - 0.4}{0.0555} \approx \frac{-0.08}{0.0555} \approx -1.44
\]
### Step 3: Find the probability using the Z-score
Now, we will calculate the probability corresponding to that Z-score.
- Using a standard normal distribution table or calculator for \( Z = -1.44 \):
\[
P(Z \leq -1.44) \approx 0.0753
\]
- Therefore, the probability of \( P(X < 0.32) \) is approximately \( 0.0753 \) or 7.53%. Consequently, the probability of \( P(X > 0.32) \) is:
\[
P(X > 0.32) = 1 - P(Z \leq -1.44) \approx 1 - 0.0753 \approx 0.9247
\]
### Conclusion
To determine if it would be unusual for more than 32% of the individuals in the sample to have high blood pressure, we typically consider an event unusual if the probability is less than 0.05 (or 5%).
Since the probability of having more than 32% of individuals with high blood pressure is approximately **0.9247** (or 92.47%), it **wouldn't** be unusual.
**Final Answer:**
It **wouldn't** be unusual since the probability is **0.9247** (or 92.47%).
### Step 1: Calculate the mean (\( \mu \)) and standard deviation (\( \sigma \))
1. **Population proportion**: The population proportion is \( p = 0.4 \).
2. **Sample size**: The sample size is \( n = 78 \).
- The mean \( \mu \) of the sampling distribution of the sample proportion is:
\[
\mu = p = 0.4
\]
- The standard deviation \( \sigma \) of the sampling distribution is calculated using the formula:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{78}} \approx \sqrt{\frac{0.24}{78}} \approx \sqrt{0.0030769} \approx 0.0555
\]
### Step 2: Convert the proportion to a Z-score
Next, convert the proportion of 32% into a Z-score.
- For \( p = 0.32 \):
\[
Z = \frac{p - \mu}{\sigma} = \frac{0.32 - 0.4}{0.0555} \approx \frac{-0.08}{0.0555} \approx -1.44
\]
### Step 3: Find the probability using the Z-score
Now, we will calculate the probability corresponding to that Z-score.
- Using a standard normal distribution table or calculator for \( Z = -1.44 \):
\[
P(Z \leq -1.44) \approx 0.0753
\]
- Therefore, the probability of \( P(X < 0.32) \) is approximately \( 0.0753 \) or 7.53%. Consequently, the probability of \( P(X > 0.32) \) is:
\[
P(X > 0.32) = 1 - P(Z \leq -1.44) \approx 1 - 0.0753 \approx 0.9247
\]
### Conclusion
To determine if it would be unusual for more than 32% of the individuals in the sample to have high blood pressure, we typically consider an event unusual if the probability is less than 0.05 (or 5%).
Since the probability of having more than 32% of individuals with high blood pressure is approximately **0.9247** (or 92.47%), it **wouldn't** be unusual.
**Final Answer:**
It **wouldn't** be unusual since the probability is **0.9247** (or 92.47%).