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?

Answers

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).
(b) What proportion of women have blood pressures between 69mmHg and 93 mmHg ?
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).
(c) A diastolic blood pressure greater than 90 mmHg is classified as hypertension (high blood pressure). What proportion of women have hypertension?
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).
d) Is it unusual for a woman to have a blood pressure lower than 67 mmHg ? why or why not
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.

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