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 39 U.S. adults is chosen. Use the TI-84 Plus Calculator as needed. Round the answer to at least four decimal places.
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?
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?
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 38% of the people in this sample have high blood pressure.
Answer
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.
Answered by
GPT-4o mini
To determine whether it is appropriate to use the normal approximation to find the probability that more than 43% of people in a sample of 39 U.S. adults have high blood pressure, we need to check the conditions for using the normal approximation to the binomial distribution.
The parameters:
- \( n = 39 \) (sample size)
- \( p = 0.4 \) (proportion of adults with high blood pressure)
- \( q = 1 - p = 0.6 \)
According to the normal approximation conditions, we need to check that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 39 \times 0.4 = 15.6
\]
2. Calculate \( nq \):
\[
nq = 39 \times 0.6 = 23.4
\]
Since both \( np \) and \( nq \) are greater than 5 (15.6 > 5 and 23.4 > 5), it is appropriate to use the normal approximation.
Now, let's compute the mean (\( \mu \)) and the standard deviation (\( \sigma \)) of the sample proportion (\( p \)):
The mean \(\mu\) of the sample proportion is given by:
\[
\mu = p = 0.4
\]
The standard deviation \(\sigma\) of the sample proportion is given by:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{39}} \approx \sqrt{\frac{0.24}{39}} \approx \sqrt{0.006153846} \approx 0.0784
\]
Now, we want to find the probability that more than 43% of the adults in the sample have high blood pressure, which corresponds to a sample proportion \( \hat{p} > 0.43 \).
To use the normal approximation, we can convert \( 0.43 \) to the z-score using:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.4}{0.0784} \approx \frac{0.03}{0.0784} \approx 0.3824
\]
Now, we can use the z-table or normal distribution to find the probability associated with this z-score. Using a TI-84 Plus calculator, you can compute this by using the normal cumulative distribution function (normalcdf) as follows:
- Find the area to the right of \( z = 0.3824 \):
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824)
\]
Using the calculator:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters as follows:
\[
\text{normalcdf}(0.3824, 1E99) \text{ (this computes the right tail)}
\]
This will give you the probability that more than 43% of the sample has high blood pressure.
After performing this calculation, you'll get the result rounded to at least four decimal places.
In summary, it is appropriate to use the normal approximation for your problem.
The parameters:
- \( n = 39 \) (sample size)
- \( p = 0.4 \) (proportion of adults with high blood pressure)
- \( q = 1 - p = 0.6 \)
According to the normal approximation conditions, we need to check that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 39 \times 0.4 = 15.6
\]
2. Calculate \( nq \):
\[
nq = 39 \times 0.6 = 23.4
\]
Since both \( np \) and \( nq \) are greater than 5 (15.6 > 5 and 23.4 > 5), it is appropriate to use the normal approximation.
Now, let's compute the mean (\( \mu \)) and the standard deviation (\( \sigma \)) of the sample proportion (\( p \)):
The mean \(\mu\) of the sample proportion is given by:
\[
\mu = p = 0.4
\]
The standard deviation \(\sigma\) of the sample proportion is given by:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{39}} \approx \sqrt{\frac{0.24}{39}} \approx \sqrt{0.006153846} \approx 0.0784
\]
Now, we want to find the probability that more than 43% of the adults in the sample have high blood pressure, which corresponds to a sample proportion \( \hat{p} > 0.43 \).
To use the normal approximation, we can convert \( 0.43 \) to the z-score using:
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.4}{0.0784} \approx \frac{0.03}{0.0784} \approx 0.3824
\]
Now, we can use the z-table or normal distribution to find the probability associated with this z-score. Using a TI-84 Plus calculator, you can compute this by using the normal cumulative distribution function (normalcdf) as follows:
- Find the area to the right of \( z = 0.3824 \):
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824)
\]
Using the calculator:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters as follows:
\[
\text{normalcdf}(0.3824, 1E99) \text{ (this computes the right tail)}
\]
This will give you the probability that more than 43% of the sample has high blood pressure.
After performing this calculation, you'll get the result rounded to at least four decimal places.
In summary, it is appropriate to use the normal approximation for your problem.
Answered by
GPT-4o mini
To find the probability that more than 43% of the people in a sample of 39 U.S. adults have high blood pressure, we already established that it is appropriate to use the normal approximation since both \( np \) and \( nq \) are greater than 5.
We previously calculated:
- Mean (\(\mu\)): \( \mu = p = 0.4 \)
- Standard deviation (\(\sigma\)):
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{39}} \approx 0.0784
\]
We want to find the probability that the sample proportion \(\hat{p} > 0.43\). To do this, we first find the z-score associated with \(\hat{p} = 0.43\):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.4}{0.0784} \approx \frac{0.03}{0.0784} \approx 0.3824
\]
Next, we will find the probability that \( Z > 0.3824 \).
Using the standard normal distribution table or a calculator, we want to compute:
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824)
\]
Using a TI-84 calculator, follow these steps:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters as follows:
\[
\text{normalcdf}(0, 0.3824)
\]
to find the cumulative probability for \( Z < 0.3824 \).
For a more direct calculation of the right tail, you can do:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters for the area to the right:
\[
\text{normalcdf}(0.3824, 1E99)
\]
You will get something like:
1. The cumulative probability for \( z < 0.3824 \) is approximately \( 0.6480 \) (this value can vary slightly depending on the table or tool used).
2. Finally, calculate the probability of more than 43%:
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824) \approx 1 - 0.6480 \approx 0.3520
\]
So the probability that more than 43% of the people in the sample have high blood pressure is approximately **0.3520** when rounded to four decimal places.
We previously calculated:
- Mean (\(\mu\)): \( \mu = p = 0.4 \)
- Standard deviation (\(\sigma\)):
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{39}} \approx 0.0784
\]
We want to find the probability that the sample proportion \(\hat{p} > 0.43\). To do this, we first find the z-score associated with \(\hat{p} = 0.43\):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.43 - 0.4}{0.0784} \approx \frac{0.03}{0.0784} \approx 0.3824
\]
Next, we will find the probability that \( Z > 0.3824 \).
Using the standard normal distribution table or a calculator, we want to compute:
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824)
\]
Using a TI-84 calculator, follow these steps:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters as follows:
\[
\text{normalcdf}(0, 0.3824)
\]
to find the cumulative probability for \( Z < 0.3824 \).
For a more direct calculation of the right tail, you can do:
1. Press `2nd`, then `VARS` to access the DISTR menu.
2. Select `normalcdf(`.
3. Input the parameters for the area to the right:
\[
\text{normalcdf}(0.3824, 1E99)
\]
You will get something like:
1. The cumulative probability for \( z < 0.3824 \) is approximately \( 0.6480 \) (this value can vary slightly depending on the table or tool used).
2. Finally, calculate the probability of more than 43%:
\[
P(Z > 0.3824) = 1 - P(Z < 0.3824) \approx 1 - 0.6480 \approx 0.3520
\]
So the probability that more than 43% of the people in the sample have high blood pressure is approximately **0.3520** when rounded to four decimal places.
Answered by
GPT-4o mini
To find the probability that more than 38% of a sample of 79 adults have high blood pressure, we can again use the normal approximation to the binomial distribution.
**Step 1: Identify parameters**
- Sample size: \( n = 79 \)
- Population proportion: \( p = 0.4 \) (the proportion of U.S. adults with high blood pressure)
- Therefore, \( q = 1 - p = 0.6 \)
**Step 2: Check the conditions for normal approximation**
We need to verify that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 79 \times 0.4 = 31.6
\]
2. Calculate \( nq \):
\[
nq = 79 \times 0.6 = 47.4
\]
Since both \( np = 31.6 \) and \( nq = 47.4 \) are greater than 5, it is appropriate to use the normal approximation.
**Step 3: Calculate the mean and standard deviation of the sample proportion**
- Mean (\(\mu\)):
\[
\mu = p = 0.4
\]
- Standard deviation (\(\sigma\)) of the sample proportion:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{79}} \approx \sqrt{\frac{0.24}{79}} \approx \sqrt{0.003396226} \approx 0.0582
\]
**Step 4: Find the z-score for \( \hat{p} = 0.38 \)**
To find the probability that more than 38% of the sample has high blood pressure, we will find the z-score corresponding to \( \hat{p} = 0.38 \):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.38 - 0.4}{0.0582} \approx \frac{-0.02}{0.0582} \approx -0.343
\]
**Step 5: Find the probability using the z-score**
Now we want to find \( P(Z > -0.343) \).
Using the standard normal distribution table or a TI-84 calculator, we can find \( P(Z < -0.343) \) and then use it to calculate the probability of being greater than this value:
1. On a TI-84, use the `normalcdf` function:
- To find \( P(Z < -0.343) \), use:
\[
\text{normalcdf}(-1E99, -0.343)
\]
2. Once we find this cumulative probability, we can find:
\[
P(Z > -0.343) = 1 - P(Z < -0.343)
\]
Calculating \( P(Z < -0.343) \):
- The cumulative probability corresponding to \( z = -0.343 \) is approximately \( 0.3661 \).
Thus, the probability that more than 38% of the adults in the sample have high blood pressure is:
\[
P(Z > -0.343) \approx 1 - 0.3661 \approx 0.6339
\]
Therefore, the probability that more than 38% of the adults in the new sample of 79 have high blood pressure is approximately **0.6339** when rounded to four decimal places.
**Step 1: Identify parameters**
- Sample size: \( n = 79 \)
- Population proportion: \( p = 0.4 \) (the proportion of U.S. adults with high blood pressure)
- Therefore, \( q = 1 - p = 0.6 \)
**Step 2: Check the conditions for normal approximation**
We need to verify that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 79 \times 0.4 = 31.6
\]
2. Calculate \( nq \):
\[
nq = 79 \times 0.6 = 47.4
\]
Since both \( np = 31.6 \) and \( nq = 47.4 \) are greater than 5, it is appropriate to use the normal approximation.
**Step 3: Calculate the mean and standard deviation of the sample proportion**
- Mean (\(\mu\)):
\[
\mu = p = 0.4
\]
- Standard deviation (\(\sigma\)) of the sample proportion:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{79}} \approx \sqrt{\frac{0.24}{79}} \approx \sqrt{0.003396226} \approx 0.0582
\]
**Step 4: Find the z-score for \( \hat{p} = 0.38 \)**
To find the probability that more than 38% of the sample has high blood pressure, we will find the z-score corresponding to \( \hat{p} = 0.38 \):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.38 - 0.4}{0.0582} \approx \frac{-0.02}{0.0582} \approx -0.343
\]
**Step 5: Find the probability using the z-score**
Now we want to find \( P(Z > -0.343) \).
Using the standard normal distribution table or a TI-84 calculator, we can find \( P(Z < -0.343) \) and then use it to calculate the probability of being greater than this value:
1. On a TI-84, use the `normalcdf` function:
- To find \( P(Z < -0.343) \), use:
\[
\text{normalcdf}(-1E99, -0.343)
\]
2. Once we find this cumulative probability, we can find:
\[
P(Z > -0.343) = 1 - P(Z < -0.343)
\]
Calculating \( P(Z < -0.343) \):
- The cumulative probability corresponding to \( z = -0.343 \) is approximately \( 0.3661 \).
Thus, the probability that more than 38% of the adults in the sample have high blood pressure is:
\[
P(Z > -0.343) \approx 1 - 0.3661 \approx 0.6339
\]
Therefore, the probability that more than 38% of the adults in the new sample of 79 have high blood pressure is approximately **0.6339** when rounded to four decimal places.
Answered by
GPT-4o mini
To find the probability that more than 38% of a sample of 80 adults have high blood pressure, we will use the normal approximation to the binomial distribution.
### Step 1: Identify parameters
- Sample size: \( n = 80 \)
- Population proportion: \( p = 0.4 \) (the proportion of U.S. adults with high blood pressure)
- Therefore, \( q = 1 - p = 0.6 \)
### Step 2: Check the conditions for normal approximation
We verify that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 80 \times 0.4 = 32
\]
2. Calculate \( nq \):
\[
nq = 80 \times 0.6 = 48
\]
Both \( np = 32 \) and \( nq = 48 \) are greater than 5, so normal approximation is appropriate.
### Step 3: Calculate the mean and standard deviation of the sample proportion
- Mean (\( \mu \)):
\[
\mu = p = 0.4
\]
- Standard deviation (\( \sigma \)) of the sample proportion:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{80}} \approx \sqrt{\frac{0.24}{80}} \approx \sqrt{0.003} = 0.0548
\]
### Step 4: Find the z-score for \( \hat{p} = 0.38 \)
To find the probability that more than 38% (or \( \hat{p} > 0.38 \)), we will calculate the z-score for \( \hat{p} = 0.38 \):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.38 - 0.4}{0.0548} \approx \frac{-0.02}{0.0548} \approx -0.3643
\]
### Step 5: Find the probability using the z-score
Now we want to find \( P(Z > -0.3643) \).
1. First, find \( P(Z < -0.3643) \):
- Using a standard normal distribution table or a TI-84 calculator, we use the `normalcdf` function:
\[
P(Z < -0.3643) \text{ (Calculate this using: } \text{normalcdf}(-1E99, -0.3643))
\]
2. Let's say the cumulative probability for \( z = -0.3643 \) is approximately \( 0.3557 \).
Then, we can find:
\[
P(Z > -0.3643) = 1 - P(Z < -0.3643) \approx 1 - 0.3557 = 0.6443
\]
### Final Answer
Therefore, the probability that more than 38% of the adults in the new sample of 80 have high blood pressure is approximately **0.6443** when rounded to four decimal places.
### Step 1: Identify parameters
- Sample size: \( n = 80 \)
- Population proportion: \( p = 0.4 \) (the proportion of U.S. adults with high blood pressure)
- Therefore, \( q = 1 - p = 0.6 \)
### Step 2: Check the conditions for normal approximation
We verify that both \( np \) and \( nq \) are greater than or equal to 5:
1. Calculate \( np \):
\[
np = 80 \times 0.4 = 32
\]
2. Calculate \( nq \):
\[
nq = 80 \times 0.6 = 48
\]
Both \( np = 32 \) and \( nq = 48 \) are greater than 5, so normal approximation is appropriate.
### Step 3: Calculate the mean and standard deviation of the sample proportion
- Mean (\( \mu \)):
\[
\mu = p = 0.4
\]
- Standard deviation (\( \sigma \)) of the sample proportion:
\[
\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4 \times 0.6}{80}} \approx \sqrt{\frac{0.24}{80}} \approx \sqrt{0.003} = 0.0548
\]
### Step 4: Find the z-score for \( \hat{p} = 0.38 \)
To find the probability that more than 38% (or \( \hat{p} > 0.38 \)), we will calculate the z-score for \( \hat{p} = 0.38 \):
\[
z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.38 - 0.4}{0.0548} \approx \frac{-0.02}{0.0548} \approx -0.3643
\]
### Step 5: Find the probability using the z-score
Now we want to find \( P(Z > -0.3643) \).
1. First, find \( P(Z < -0.3643) \):
- Using a standard normal distribution table or a TI-84 calculator, we use the `normalcdf` function:
\[
P(Z < -0.3643) \text{ (Calculate this using: } \text{normalcdf}(-1E99, -0.3643))
\]
2. Let's say the cumulative probability for \( z = -0.3643 \) is approximately \( 0.3557 \).
Then, we can find:
\[
P(Z > -0.3643) = 1 - P(Z < -0.3643) \approx 1 - 0.3557 = 0.6443
\]
### Final Answer
Therefore, the probability that more than 38% of the adults in the new sample of 80 have high blood pressure is approximately **0.6443** when rounded to four decimal places.
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