Answer:
The percentage of individuals from this population will have LDL levels 1 or more standard deviations above the mean is 16%.
Step-by-step explanation:
Let X represent the LDL levels.
It is provided that [tex]X\sim N(123, 41^{2})[/tex].
Compute the probability that a randomly selected individual is will have LDL levels 1 or more standard deviations above the mean as follows:
[tex]P(X\geq \mu+\sigma)=P(X\geq 123+41)[/tex]
           [tex]=P(X\geq 164)\\\\=P(\frac{X-\mu}{\sigma}\geq \frac{164-123}{41})\\\\=P(Z>1)\\\\=1-P(Z<1)\\\\=1-0.84134\\\\=0.15866\\\\\approx 0.16[/tex]
Thus, the percentage of individuals from this population will have LDL levels 1 or more standard deviations above the mean is 16%.