Question

The mean cholesterol level of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l. Assume that cholesterol levels are normally distributed. Round the probabilities to four decimal places. It is possible with rounding for a probability to be 0.0000.

a) State the random variable. rv X = the cholesterol level of a randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles Correct

b) Find the probability that a randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level of 3.5 mmol/l or more.

c) Find the probability that a randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level of 8.1 mmol/l or less.

d) Find the probability that a randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level between 3.5 and 8.1 mmol/l.

e) Find the probability that randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles has a cholesterol level that is at least 8.6 mmol/l.

f) Is a cholesterol level of 8.6 mmol/l unusually high for a randomly selected woman age 45-59 in Ghana, Nigeria, or Seychelles? Why or why not? Select an answer

g) What cholesterol level do 67% of all women age 45-59 in Ghana, Nigeria, and Seychelles have less than? Round your answer to two decimal places in the first box. Put the correct units in the second box.

Answer 1

Answer :

Given in the question

Mean = 5.1

Standard deviation = 1

Solution(a)

Cholesterol levels of women

Solution(b)

From Z table we found p-value

= 0.945201

Solution(c)

= 3

From Z table we found p-value

Solution(d)

= 0.8504

Solution(e)

= 3.3

From Z table we found p-value

Solution(f)

Yes, this is unusually high because Z score is more than 2.

Solution(g)

P-value = 0.67

Z-score = -0.43991

-0.43991= (X-5.1)/1

X = 5.1-0.43991 = 4.66

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