Question 1186331: The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l (Lawes, Hoorn, Law & Rodgers, 2004). Assume that cholesterol levels are normally distributed.
a.) State the random variable.
b.) Find the probability that a woman age 45-59 in Ghana has a cholesterol level above 6.2 mmol/l (considered a high level).
c.) Suppose doctors decide to test the woman’s cholesterol level again and average the two values. Find the probability that this woman’s mean cholesterol level for the two tests is above 6.2 mmol/l.
d.) Suppose doctors being very conservative decide to test the woman’s cholesterol level a third time and average the three values. Find the probability that this woman’s mean cholesterol level for the three tests is above 6.2 mmol/l.
e.) If the sample mean cholesterol level for this woman after three tests is above 6.2 mmol/l, what could you conclude?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**a.) State the random variable.**
The random variable is the cholesterol level (in mmol/l) of a woman aged 45-59 in Ghana.
**b.) Probability of cholesterol level above 6.2 mmol/l (one test):**
1. **Calculate the z-score:**
z = (x - μ) / σ
z = (6.2 - 5.1) / 1.0
z = 1.1
2. **Find the probability:**
Using a z-table or calculator, find the probability of a z-score being *greater* than 1.1. This represents the area to the *right* of 1.1 on the standard normal distribution.
P(z > 1.1) ≈ 0.1357
Therefore, the probability that a woman's cholesterol level is above 6.2 mmol/l is approximately **0.1357**.
**c.) Probability of mean cholesterol level above 6.2 mmol/l (two tests):**
When averaging two tests, we are dealing with the *sampling distribution of the mean*. The standard deviation of this distribution (also called the standard error) is:
Standard Error = σ / √n = 1.0 / √2 ≈ 0.7071
1. **Calculate the z-score:**
z = (x̄ - μ) / (σ / √n)
z = (6.2 - 5.1) / 0.7071
z ≈ 1.56
2. **Find the probability:**
P(z > 1.56) ≈ 0.0594
Therefore, the probability of the mean cholesterol level for two tests being above 6.2 mmol/l is approximately **0.0594**.
**d.) Probability of mean cholesterol level above 6.2 mmol/l (three tests):**
The standard error for three tests is:
Standard Error = σ / √n = 1.0 / √3 ≈ 0.5774
1. **Calculate the z-score:**
z = (x̄ - μ) / (σ / √n)
z = (6.2 - 5.1) / 0.5774
z ≈ 1.90
2. **Find the probability:**
P(z > 1.90) ≈ 0.0287
Therefore, the probability of the mean cholesterol level for three tests being above 6.2 mmol/l is approximately **0.0287**.
**e.) Conclusion if the sample mean after three tests is above 6.2 mmol/l:**
If the sample mean cholesterol level after three tests is above 6.2 mmol/l, we have stronger evidence to suggest that the woman's *true* mean cholesterol level may be higher than 5.1 mmol/l. While there is still a small probability (about 2.87%) of observing this result if the true mean is actually 5.1, the repeated measurements make it less likely that it is due to random chance.
It's important to remember that this is still a probability. We can't definitively say her mean is higher, but the evidence suggests it is more probable than before the repeated tests. Further testing or investigation might be warranted.
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