Answer:
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
Explanation:
given data
Normal distribution
mean cholesterol level μ= 51.6 mg/dL
Standard deviation σ= 14.3 mg/dL
x= 60 mg/dL
We have to find out P(x>60)
We Know that [tex]P(x>a) =P(Z>(a-\mu)/\sigma)[/tex]
therefore, [tex]P(x>60) =P(Z>(60-51.3)/14.3)[/tex]
= P(Z>0.61)
= 1 - P(Z<0.61)
= 1 - 0.7291
= 0.27
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
