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.

Question

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.

Answer 1

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: \[ \boxed{0.0582} \]

Answer 2

Find the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between .24 and .35.

Answer 3

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:

\[ \boxed{0.7151} \]

Answer 4

Find the probability that less than 26% of the people in the sample of 80 have high blood pressure.

Answer 5

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:

\[ \boxed{0.2181} \]

Answer 6

Would it be unusual if more than42% of the individuals in the sample of 80 had high blood pressure?

Answer 7

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.

\[ \text{Final Conclusion: } \text{Yes, it would be unusual.} \]