The mean serum cholesterol level for U.S. adults was 204, with a standard deviation of 42.5 (the units are milligrams per deciliter). A simple random sample of 112 adults is chosen. Round the answers to four decimal places.

Question

What is the probability that the sample mean cholesterol level is greater than 212?

Answer 1

To find the probability that the sample mean cholesterol level is greater than 212 mg/dL, we can use the Central Limit Theorem. According to this theorem, if the sample size is large enough (which it is, in this case), the sampling distribution of the sample mean will be approximately normally distributed.

Here are the steps to solve the problem:

1. Calculate the Standard Error of the Mean (SEM)

The Standard Error of the Mean can be calculated using the formula:

\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]

where:

• \(\sigma\) = standard deviation of the population = 42.5 mg/dL
• \(n\) = sample size = 112

Calculating the SEM:

\[ \text{SEM} = \frac{42.5}{\sqrt{112}} \approx \frac{42.5}{10.583} \approx 4.0187 \]

2. Calculate the Z-score for the sample mean of 212 mg/dL

Now we calculate the Z-score using the formula:

\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \]

where:

• \(\bar{x}\) = sample mean = 212 mg/dL
• \(\mu\) = population mean = 204 mg/dL

Calculating the Z-score:

\[ Z = \frac{212 - 204}{4.0187} \approx \frac{8}{4.0187} \approx 1.9873 \]

3. Find the Probability

Now we need to find the probability that the sample mean is greater than 212 mg/dL, which corresponds to finding \(P(Z > 1.9873)\).

Using standard normal distribution tables or a calculator:

\[ P(Z > 1.9873) = 1 - P(Z < 1.9873) \]

From Z-tables or using a calculator, we find:

\[ P(Z < 1.9873) \approx 0.9761 \]

Thus,

\[ P(Z > 1.9873) = 1 - 0.9761 = 0.0239 \]

Conclusion

The probability that the sample mean cholesterol level is greater than 212 mg/dL is approximately:

\[ \boxed{0.0239} \]

Answer 2

What is the probability that the sample mean cholesterol level is between 192 and 202?

Answer 3

To find the probability that the sample mean cholesterol level is between 192 mg/dL and 202 mg/dL, we can again use the Central Limit Theorem and the properties of the normal distribution.

Step 1: Calculate the Standard Error of the Mean (SEM)

As calculated previously, the Standard Error of the Mean (SEM) is:

\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{42.5}{\sqrt{112}} \approx 4.0187. \]

Step 2: Calculate the Z-scores for 192 mg/dL and 202 mg/dL

Z-score for 192 mg/dL

\[ Z_{192} = \frac{192 - 204}{\text{SEM}} = \frac{192 - 204}{4.0187} = \frac{-12}{4.0187} \approx -2.98. \]

Z-score for 202 mg/dL

\[ Z_{202} = \frac{202 - 204}{\text{SEM}} = \frac{202 - 204}{4.0187} = \frac{-2}{4.0187} \approx -0.4977. \]

Step 3: Find the Probabilities for Each Z-score

Using standard normal distribution tables or a calculator:

1. For \(Z_{192} \approx -2.98\):

   \[ P(Z < -2.98) \approx 0.0014. \]

2. For \(Z_{202} \approx -0.4977\):

   \[ P(Z < -0.4977) \approx 0.3106. \]

Step 4: Find the Probability Between the Two Z-scores

We want to find the probability that \(Z\) is between \(-2.98\) and \(-0.4977\):

\[ P(-2.98 < Z < -0.4977) = P(Z < -0.4977) - P(Z < -2.98). \]

Substituting the probabilities:

\[ P(-2.98 < Z < -0.4977) = 0.3106 - 0.0014 = 0.3092. \]

Conclusion

The probability that the sample mean cholesterol level is between 192 mg/dL and 202 mg/dL is approximately:

\[ \boxed{0.3092}. \]

Answer 4

Would it be unusual for the sample mean to be less than 197?

Answer 5

To determine whether it would be unusual for the sample mean cholesterol level to be less than 197 mg/dL, we can calculate the Z-score for 197 mg/dL and find the corresponding probability.

Step 1: Calculate the Z-score for 197 mg/dL

Using the formula for the Z-score:

\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \]

where:

• \(\bar{x} = 197\) mg/dL (the sample mean we are considering),
• \(\mu = 204\) mg/dL (the population mean),
• \(\text{SEM} = 4.0187\) (calculated previously).

Substituting the values:

\[ Z = \frac{197 - 204}{4.0187} = \frac{-7}{4.0187} \approx -1.738. \]

Step 2: Find the Probability for the Z-score

Now we find the probability associated with this Z-score using standard normal distribution tables or a calculator:

\[ P(Z < -1.738). \]

Using a standard normal distribution table or calculator, we find:

\[ P(Z < -1.738) \approx 0.0415. \]

Step 3: Interpret the Probability

A probability of \(0.0415\) (or \(4.15%\)) indicates that there is a relatively low chance (about 4.15%) of obtaining a sample mean less than 197 mg/dL.

In statistical terms, if we define unusual events as those with probabilities less than \(0.05\) (5%), then:

• The sample mean being less than \(197\) mg/dL is considered unusual.

Conclusion

Yes, it would be unusual for the sample mean cholesterol level to be less than \(197\) mg/dL, as evidenced by the probability of approximately \(0.0415\). Thus, we can conclude:

\[ \text{It is unusual for the sample mean to be less than } 197 \text{ mg/dL}. \]