The mean serum cholesterol level for U.S. adults was 202, with a standard deviation of 41.6 (the units are milligrams per deciliter). A simple random sample of 112 adults is chosen.

Question

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

What is the probability that the sample mean cholesterol level is between 191 and 199?

Would it be unusual for the sample mean to be less than 199? Why or why not

Answer 1

To assess the probabilities regarding the sample mean cholesterol levels, we can use the Central Limit Theorem. According to this theorem, the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (which it is in this case).

1. Determine the mean and standard deviation of the sample mean:

   • The mean of the sampling distribution (the population mean) is \( \mu = 202 \).
   • The standard deviation of the sampling distribution, known as the standard error (SE), is given by:

   \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{41.6}{\sqrt{112}} \approx \frac{41.6}{10.583} \approx 3.93 \]

2. Probability that the sample mean cholesterol level is greater than 212:

   • To find this probability, we can standardize the sample mean using the Z-score formula:

   \[ Z = \frac{X - \mu}{SE} = \frac{212 - 202}{3.93} \approx \frac{10}{3.93} \approx 2.54 \]

   • Now, we look up this Z-score in the standard normal distribution table, or use a calculator to find:

   \[ P(Z > 2.54) \approx 1 - P(Z \leq 2.54) \approx 1 - 0.9945 = 0.0055 \]

   Therefore, the probability that the sample mean is greater than 212 is approximately 0.0055, or 0.55%.

3. Probability that the sample mean cholesterol level is between 191 and 199:

   • We will compute the Z-scores for both 191 and 199.

   For 191: \[ Z_{191} = \frac{191 - 202}{3.93} \approx \frac{-11}{3.93} \approx -2.80 \]

   For 199: \[ Z_{199} = \frac{199 - 202}{3.93} \approx \frac{-3}{3.93} \approx -0.76 \]

   • Now, using the Z-table:

   \[ P(Z \leq -2.80) \approx 0.0026 \] \[ P(Z \leq -0.76) \approx 0.2236 \]

   • Therefore, the probability that the sample mean is between 191 and 199 is:

   \[ P(-2.80 < Z < -0.76) = P(Z \leq -0.76) - P(Z \leq -2.80) \approx 0.2236 - 0.0026 \approx 0.2210 \]

   So, the probability that the sample mean is between 191 and 199 is approximately 0.2210, or 22.10%.

4. Would it be unusual for the sample mean to be less than 199?

   • To determine if it's unusual for the sample mean to be less than 199, we can compute the probability:

   \[ Z_{199} = \frac{199 - 202}{3.93} \approx -0.76 \]

   • From the Z-table, we found that \( P(Z < -0.76) \approx 0.2236 \), which indicates that approximately 22.36% of the time the sample mean would be less than 199.

   • Generally, thresholds of 5% (0.05) are considered unusual in statistics. Since 22.36% is considerably higher than 5%, it would not be unusual for the sample mean to be less than 199.

Answer 2

What is the probability that the sample mean cholesterol level is between 191 and 199? the answer is not .2210

Answer 3

Let's carefully recalculate the probability that the sample mean cholesterol level is between 191 and 199.

We already know:

• The population mean (\( \mu \)) is 202.
• The standard deviation (\( \sigma \)) is 41.6.
• The sample size (\( n \)) is 112.
• The standard error (SE) of the sample mean is approximately 3.93.

We need to calculate the Z-scores for 191 and 199:

1. Z-score for 191:

\[ Z_{191} = \frac{191 - 202}{SE} = \frac{191 - 202}{3.93} \approx \frac{-11}{3.93} \approx -2.80 \]

2. Z-score for 199:

\[ Z_{199} = \frac{199 - 202}{SE} = \frac{199 - 202}{3.93} \approx \frac{-3}{3.93} \approx -0.76 \]

3. Find the probabilities from the standard normal distribution:

Using the Z-scores:

• For \( Z = -2.80 \):

Using a Z-table or calculator:

\[ P(Z < -2.80) \approx 0.0026 \]

• For \( Z = -0.76 \):

\[ P(Z < -0.76) \approx 0.2236 \]

4. Calculate the probability between 191 and 199:

To find the probability that the sample mean is between 191 and 199:

\[ P(191 < \bar{X} < 199) = P(Z < -0.76) - P(Z < -2.80) \]

Substituting the values:

\[ P(191 < \bar{X} < 199) \approx 0.2236 - 0.0026 = 0.2210 \]

Thus, the calculation holds, and the probability that the sample mean cholesterol level is between 191 and 199 is indeed approximately 0.2210 or 22.10%.

If you have a specific answer or calculation in mind that you believe is correct, please provide it so I can address any discrepancies!

Answer 4

Would it be unusual for the sample mean to be less than 199? why or why not

Answer 5

To determine whether it would be unusual for the sample mean cholesterol level to be less than 199, we need to consider the probability of this event occurring.

Calculation Summary:

1. Mean of Sample Mean (μ): 202 mg/dL

2. Standard Error (SE): Approximately 3.93 mg/dL

3. Z-score for Sample Mean < 199:

   \[ Z_{199} = \frac{199 - 202}{SE} = \frac{199 - 202}{3.93} \approx \frac{-3}{3.93} \approx -0.76 \]

4. Probability for Z-score:

   • Using the standard normal distribution, we find:

   \[ P(Z < -0.76) \approx 0.2236 \]

Interpretation of the Probability:

• The probability \( P(\bar{X} < 199) \approx 0.2236 \) means that approximately 22.36% of the time, a simple random sample of 112 adults will yield a mean cholesterol level less than 199 mg/dL.

Conclusion:

• In statistical terms, an event is often considered "unusual" if it has a probability of occurring that is less than 5% (0.05). Since the probability of obtaining a sample mean less than 199 is approximately 22.36%, this is not below the threshold of 5%.

Final Answer:

• Conclusion: It would not be unusual for the sample mean to be less than 199 mg/dL because the probability of this happening is relatively high (22.36%), well above the typical threshold of 5% for unusual events.