Assuming that the distribution of serum cholesterol level in men is Normal with mean 242.2 and standard deviation 45.4, calculate the proportion of men with serum cholesterol levels in the interval 187.7 to 328.5.
Accepted Solution
A:
To find the proportion of men with serum cholesterol levels in the interval 187.7 to 328.5, we need to calculate the area under the normal curve between these two values.
Let's denote the mean as $\mu = 242.2$ and the standard deviation as $\sigma = 45.4$.
To standardize the interval, we calculate the z-scores for the lower and upper boundaries: $ z_1 = \frac{{187.7 - \mu}}{{\sigma}}$ and $z_2 = \frac{{328.5 - \mu}}{{\sigma}}$.
Substituting the given values, we have $z_1 = \frac{{187.7 - 242.2}}{{45.4}} \approx -1.20$ and $z_2 = \frac{{328.5 - 242.2}}{{45.4}} \approx 1.90$.
Using a standard normal distribution table or calculator, we can find the proportion of values falling between these two z-scores.
The proportion is given by: $$P( -1.20 \leq z \leq 1.90)$$
By looking up the z-scores in the standard normal distribution table,
Since we want the proportion between these two z-scores, we subtract the smaller probability from the larger probability: $$P( -1.20 \leq z \leq 1.90) \approx 0.8564$$
Therefore, the proportion of men with serum cholesterol levels in the interval 187.7 to 328.5 is approximately 0.8562.