The mean serum cholesterol level for U.S. adults was 203, with a standard deviation of 42 (the units are milligrams per deciliter). A simple random sample of 109 adults is chosen. Use the TI-84 calculator. Round the answers to foi decimal places Part 1 of 3 What is the probability that the sample mean cholesterol level is greater than 213? The probability that the sample mean cholesterol level is greater than 213 is . Part 2 of 3 What is the probability that the sample mean cholesterol level is between 191 and 2012 The probability that the sample mean cholesterol level is between 191 and 201 is Part 3 of 3 Would it be unusual for the sample mean to be less than 191? be unusual for the sample mean to be less than 191, since the probability It (Choose one) is
The Correct Answer and Explanation is :
Part 1 of 3: Probability that the sample mean is greater than 213
We are given:
- The population mean cholesterol level: ( \mu = 203 ) mg/dL
- The population standard deviation: ( \sigma = 42 ) mg/dL
- Sample size: ( n = 109 )
We can use the sampling distribution of the sample mean to calculate probabilities. The sampling distribution of the sample mean has:
- A mean ( \mu_{\bar{x}} = \mu = 203 ) mg/dL
- A standard error ( SE = \frac{\sigma}{\sqrt{n}} )
The standard error is calculated as:
[
SE = \frac{42}{\sqrt{109}} \approx 4.007
]
Now, we want to find the probability that the sample mean is greater than 213 mg/dL. This corresponds to the Z-score calculation for 213:
[
Z = \frac{X – \mu_{\bar{x}}}{SE} = \frac{213 – 203}{4.007} \approx 2.495
]
We use a Z-table or the TI-84 calculator to find the area to the right of a Z-score of 2.495. Using the TI-84 calculator, we find:
[
P(Z > 2.495) \approx 0.0063
]
Thus, the probability that the sample mean is greater than 213 mg/dL is approximately 0.0063.
Part 2 of 3: Probability that the sample mean is between 191 and 201
We want to find the probability that the sample mean is between 191 and 201 mg/dL. We will calculate the Z-scores for both values:
- For ( X = 191 ):
[
Z = \frac{191 – 203}{4.007} \approx -3.000
]
- For ( X = 201 ):
[
Z = \frac{201 – 203}{4.007} \approx -0.499
]
Now, we find the probabilities associated with these Z-scores:
[
P(-3.000 < Z < -0.499)
]
Using the TI-84, we find the areas corresponding to these Z-scores:
- ( P(Z < -3.000) \approx 0.0013 )
- ( P(Z < -0.499) \approx 0.3106 )
The probability that the sample mean is between 191 and 201 is:
[
P(191 < \bar{x} < 201) = P(Z < -0.499) – P(Z < -3.000) \approx 0.3106 – 0.0013 = 0.3093
]
Thus, the probability that the sample mean is between 191 and 201 mg/dL is approximately 0.3093.
Part 3 of 3: Would it be unusual for the sample mean to be less than 191?
To determine if it would be unusual for the sample mean to be less than 191, we need to calculate the Z-score for 191 mg/dL:
[
Z = \frac{191 – 203}{4.007} \approx -3.000
]
A Z-score of -3.000 corresponds to a very small probability (about 0.0013). This is extremely unlikely in a normal distribution, as a probability of less than 0.05 (or 5%) is considered unusual.
Thus, it would be unusual for the sample mean to be less than 191 mg/dL, since the probability is very small.
Summary:
- The probability that the sample mean cholesterol level is greater than 213 is approximately 0.0063.
- The probability that the sample mean cholesterol level is between 191 and 201 is approximately 0.3093.
- It would be unusual for the sample mean to be less than 191, since the probability is about 0.0013.