Question
In a recent study, the Centers for Disease Control and prevention reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.7mmHg and a standard deviation of 10.1mmHg. Round the answers to four decimal places.
(a) What proportion of women have blood pressures lower than 68 mmHg?
(a) What proportion of women have blood pressures lower than 68 mmHg?
Answers
Answer
(b) What proportion of women have blood pressures between 69mmHg and 93 mmHg ?
Answer
(c) A diastolic blood pressure greater than 90 mmHg is classified as hypertension (high blood pressure). What proportion of women have hypertension?
Answer
d) Is it unusual for a woman to have a blood pressure lower than 67 mmHg ? why or why not
Answered by
GPT-4o mini
To find the proportion of women with diastolic blood pressures lower than 68 mmHg, we can use the properties of the normal distribution.
First, we need to calculate the z-score using the following formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value for which we want to find the proportion (68 mmHg),
- \(\mu\) is the mean (80.7 mmHg),
- \(\sigma\) is the standard deviation (10.1 mmHg).
Plugging in the numbers:
\[
z = \frac{(68 - 80.7)}{10.1} = \frac{-12.7}{10.1} \approx -1.2545
\]
Next, we use the z-score to find the corresponding proportion in a standard normal distribution table or using a calculator with normal distribution functions.
Looking up the z-score of -1.2545 (or using a calculator), we find:
\[
P(Z < -1.2545) \approx 0.1056
\]
Therefore, the proportion of women with diastolic blood pressures lower than 68 mmHg is approximately **0.1056** (rounded to four decimal places).
First, we need to calculate the z-score using the following formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the value for which we want to find the proportion (68 mmHg),
- \(\mu\) is the mean (80.7 mmHg),
- \(\sigma\) is the standard deviation (10.1 mmHg).
Plugging in the numbers:
\[
z = \frac{(68 - 80.7)}{10.1} = \frac{-12.7}{10.1} \approx -1.2545
\]
Next, we use the z-score to find the corresponding proportion in a standard normal distribution table or using a calculator with normal distribution functions.
Looking up the z-score of -1.2545 (or using a calculator), we find:
\[
P(Z < -1.2545) \approx 0.1056
\]
Therefore, the proportion of women with diastolic blood pressures lower than 68 mmHg is approximately **0.1056** (rounded to four decimal places).
Answered by
GPT-4o mini
To find the proportion of women with diastolic blood pressures between 69 mmHg and 93 mmHg, we will calculate the z-scores for both values and then find the corresponding probabilities.
### Step 1: Calculate the z-scores
1. **For 69 mmHg:**
\[
z_1 = \frac{(X - \mu)}{\sigma} = \frac{(69 - 80.7)}{10.1} = \frac{-11.7}{10.1} \approx -1.1574
\]
2. **For 93 mmHg:**
\[
z_2 = \frac{(X - \mu)}{\sigma} = \frac{(93 - 80.7)}{10.1} = \frac{12.3}{10.1} \approx 1.2178
\]
### Step 2: Find the corresponding probabilities
Now, we look up the z-scores in a standard normal distribution table or use a calculator.
1. For \(z_1 \approx -1.1574\):
\[
P(Z < -1.1574) \approx 0.1244
\]
2. For \(z_2 \approx 1.2178\):
\[
P(Z < 1.2178) \approx 0.8888
\]
### Step 3: Calculate the proportion between the two z-scores
To find the proportion of women with blood pressures between 69 mmHg and 93 mmHg, we subtract the lower probability from the upper probability:
\[
P(69 < X < 93) = P(Z < 1.2178) - P(Z < -1.1574) = 0.8888 - 0.1244 = 0.7644
\]
### Final Answer
Thus, the proportion of women with diastolic blood pressures between 69 mmHg and 93 mmHg is approximately **0.7644** (rounded to four decimal places).
### Step 1: Calculate the z-scores
1. **For 69 mmHg:**
\[
z_1 = \frac{(X - \mu)}{\sigma} = \frac{(69 - 80.7)}{10.1} = \frac{-11.7}{10.1} \approx -1.1574
\]
2. **For 93 mmHg:**
\[
z_2 = \frac{(X - \mu)}{\sigma} = \frac{(93 - 80.7)}{10.1} = \frac{12.3}{10.1} \approx 1.2178
\]
### Step 2: Find the corresponding probabilities
Now, we look up the z-scores in a standard normal distribution table or use a calculator.
1. For \(z_1 \approx -1.1574\):
\[
P(Z < -1.1574) \approx 0.1244
\]
2. For \(z_2 \approx 1.2178\):
\[
P(Z < 1.2178) \approx 0.8888
\]
### Step 3: Calculate the proportion between the two z-scores
To find the proportion of women with blood pressures between 69 mmHg and 93 mmHg, we subtract the lower probability from the upper probability:
\[
P(69 < X < 93) = P(Z < 1.2178) - P(Z < -1.1574) = 0.8888 - 0.1244 = 0.7644
\]
### Final Answer
Thus, the proportion of women with diastolic blood pressures between 69 mmHg and 93 mmHg is approximately **0.7644** (rounded to four decimal places).
Answered by
GPT-4o mini
To find the proportion of women with diastolic blood pressures greater than 90 mmHg, we will calculate the z-score for 90 mmHg and then find the corresponding probability.
### Step 1: Calculate the z-score
Using the formula for the z-score:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X = 90\) mmHg,
- \(\mu = 80.7\) mmHg (mean),
- \(\sigma = 10.1\) mmHg (standard deviation).
Calculating the z-score:
\[
z = \frac{(90 - 80.7)}{10.1} = \frac{9.3}{10.1} \approx 0.9208
\]
### Step 2: Find the probability for the z-score
Next, we find \(P(Z < 0.9208)\) using a standard normal distribution table or a calculator:
\[
P(Z < 0.9208) \approx 0.8212
\]
### Step 3: Calculate the proportion greater than 90 mmHg
To find the proportion of women with blood pressures greater than 90 mmHg, we calculate:
\[
P(X > 90) = 1 - P(Z < 0.9208) = 1 - 0.8212 = 0.1788
\]
### Final Answer
Thus, the proportion of women classified as having hypertension (diastolic blood pressure greater than 90 mmHg) is approximately **0.1788** (rounded to four decimal places).
### Step 1: Calculate the z-score
Using the formula for the z-score:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X = 90\) mmHg,
- \(\mu = 80.7\) mmHg (mean),
- \(\sigma = 10.1\) mmHg (standard deviation).
Calculating the z-score:
\[
z = \frac{(90 - 80.7)}{10.1} = \frac{9.3}{10.1} \approx 0.9208
\]
### Step 2: Find the probability for the z-score
Next, we find \(P(Z < 0.9208)\) using a standard normal distribution table or a calculator:
\[
P(Z < 0.9208) \approx 0.8212
\]
### Step 3: Calculate the proportion greater than 90 mmHg
To find the proportion of women with blood pressures greater than 90 mmHg, we calculate:
\[
P(X > 90) = 1 - P(Z < 0.9208) = 1 - 0.8212 = 0.1788
\]
### Final Answer
Thus, the proportion of women classified as having hypertension (diastolic blood pressure greater than 90 mmHg) is approximately **0.1788** (rounded to four decimal places).
Answered by
GPT-4o mini
To determine whether it is unusual for a woman to have a diastolic blood pressure lower than 67 mmHg, we can calculate the z-score for 67 mmHg and analyze its value to see how many standard deviations it is from the mean.
### Step 1: Calculate the z-score for 67 mmHg
Using the z-score formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X = 67\) mmHg,
- \(\mu = 80.7\) mmHg (mean),
- \(\sigma = 10.1\) mmHg (standard deviation).
Calculating the z-score:
\[
z = \frac{(67 - 80.7)}{10.1} = \frac{-13.7}{10.1} \approx -1.3564
\]
### Step 2: Determine probability based on the z-score
Now, we can find the probability associated with this z-score. We look up \(z = -1.3564\) in a standard normal distribution table or use a calculator.
\[
P(Z < -1.3564) \approx 0.0871
\]
### Step 3: Interpretation
The probability of a woman having a diastolic blood pressure lower than 67 mmHg is approximately **0.0871**, which means that about **8.71%** of women have a diastolic blood pressure below this level.
### Is it unusual?
In general statistical practice, values that fall below 5% or above 95% of a distribution are often considered "unusual." Since 8.71% falls above this threshold, it is not classified as unusual, although it is on the lower end of the distribution.
### Conclusion
It is not considered unusual for a woman to have a diastolic blood pressure lower than 67 mmHg, as about 8.71% fall into this category. However, it is relatively uncommon, indicating that such blood pressure levels are less frequent than more typical values.
### Step 1: Calculate the z-score for 67 mmHg
Using the z-score formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X = 67\) mmHg,
- \(\mu = 80.7\) mmHg (mean),
- \(\sigma = 10.1\) mmHg (standard deviation).
Calculating the z-score:
\[
z = \frac{(67 - 80.7)}{10.1} = \frac{-13.7}{10.1} \approx -1.3564
\]
### Step 2: Determine probability based on the z-score
Now, we can find the probability associated with this z-score. We look up \(z = -1.3564\) in a standard normal distribution table or use a calculator.
\[
P(Z < -1.3564) \approx 0.0871
\]
### Step 3: Interpretation
The probability of a woman having a diastolic blood pressure lower than 67 mmHg is approximately **0.0871**, which means that about **8.71%** of women have a diastolic blood pressure below this level.
### Is it unusual?
In general statistical practice, values that fall below 5% or above 95% of a distribution are often considered "unusual." Since 8.71% falls above this threshold, it is not classified as unusual, although it is on the lower end of the distribution.
### Conclusion
It is not considered unusual for a woman to have a diastolic blood pressure lower than 67 mmHg, as about 8.71% fall into this category. However, it is relatively uncommon, indicating that such blood pressure levels are less frequent than more typical values.