Question

Assume that the total cholesterol levels for adults are normally distributed with mean cholesterol level of 51.6 ​mg/dL and standard deviation 14.3 ​mg/dL. Find the probability that an individual will have a cholesterol level greater than 60 ​mg/dL.

Answer 1

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 $P(x>a) =P(Z>(a-\mu)/\sigma)$

therefore, $P(x>60) =P(Z>(60-51.3)/14.3)$

= 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