Question

The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/1 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 mmo/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 mmol1 e.) If the sample mean cholesterol level for this woman after three tests is above 6.2 mmol/M, what could you conclude?

Answer 1

Answer:

a) $N(\mu = 5.1 \, , \, \sigma = 1)$

b) The probability that a woman cholesterol is above 6.2 mmol/l is 0.1357

c) The probability that the sample mean cholesterol level of two tests is above 6.2 mmol/l is 0.0594

d) For three tests, the probability that the sample mean cholesterol level  is above 6.2 mmol/l is 0.0281

e) We can conclude with 95% confidence that the woman has a high level of cholesterol level.

Step-by-step explanation:

a) The random variable is $N(\mu = 5.1 \, , \, \sigma = 1)$ .

b) Lets call X the random variable. We will standarize X to obtain a standard normal random variable W

$W = \frac{X - \mu}{\sigma} = X-5.1 \approx N(0,1)$

The values of the cummulative distribution function of the standard normal random variable W, which we will denote by $\phi$ , are tabulated and they can be found in the attached file. We will use this values to find the probability of X being greater than 6.2

$P(X > 6.2) = P(X-5.1 > 6.2-5.1) = P(W > 1.1) ) 1- \phi(1.1) = 1-0.8643 = 0.1357$

Thus, the probability that a woman cholesterol is above 6.2 mmol/l is 0.1357.

c) If we take the mean of two normal random variables X1, X2 independent and with equal mean and standard deviation, then we will obtain a Normal random variable Z with the same mean (5.1) and the standard deviation is obtained by taking the standard deviation of any of the two variables divided by the square root of the sample length, in other words, by √2. This means that the mean distribution Z will have distribution $N(\mu = 5.1, \sigma = 1/\sqrt{2})$ .

We standarize Z to obtain (another) W as before,

$W = \frac{Z - 5.1}{1/\sqrt{2}} \approx N(0,1)$

And now we compute the probability of Z being greater than 1/√2 using the cummulative function $\phi$

$P(Z > 6.2) = P(W > \frac{6.2-5.1}{1/\sqrt{2}}) = P(W > 1.56) = 1-\phi(1.56)\\ = 1-0.9406 = 0.0594$

Thus, the probability that the sample mean cholesterol level is above 6.2 mmol/l is 0.0594.

d) For three values we will have a sample mean Z with mean 5.1 as before and standard deviation $\sigma = 1/\sqrt{3}$ . The standarization therefore is

$W = \frac{Z-5.1}{1/\sqrt{3}}) \approx N(0,1)$

And the probability of Z being greater than 6.2 in this case is

$P(Z > 6.2) = P(W > \frac{6.2-5.1}{1/\sqrt{3}}) = P(W > 1.91) = 1-\phi(1.91)\\ = 1-0.9719 = 0.0281$

As a result, the probability that the sample mean is greater than 6.2 mmol/l is 0.0281.

e) The sample mean being greater than 6.2 mmol/l only happens with probability 0.0281. This means that 97.19% of the woman will have a cholesterol level below that amount, and as a consecuence, this woman has a high level of cholesterol.

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