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Conditional Probability: Population Obesity and Teenage Pregnancies

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1.a) Overweight? Choose an American adult at random. Define two events:

A= the person chosen is obese
B= the person chosen is overweight but not obese

According to the 2007 National Health Interview Survey, P(A)=0.26 and P(B)=0.35. Explain why events A and B are disjoint.

1. b) Say in plain language what the event "A or B" is. What is P(AorB)?

1. c) If C is the event that the person chosen has normal weight or less, what is P(C)?

2. Older Women. Government data show that 6% of the American population are at least 75 years of age and that about 52% of Americans are women. Explain why it is wrong to conclude that because (0.06)(0.52)=0.0312 about 3% of the population are women aged 75 or older.

3.a) Teenage pregnancies. The U.S. National Center for Health Statistics reports for 2006 the distribution of teenage pregnancies by age range and ethnicity of the mother.
15 to 17 years old 18 to 19 years old
White 0.22 0.49
Black 0.09 0.17
Other 0.01 0.02

Overall, what percent of teenage pregnancies are in young mothers 15 to 17 years old?

3.b) What is the conditional probability that a teenage pregnancy is in a young mother 15 to 17 years old, given that she si classified as "white"?

3.c) What is the conditional probability that a teenage pregnancy is in a young mother 15 to 17 years old, given that she is classified as "black"?

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Solution Summary

Conditional probability for population obesity and teenage pregnancies are determined.

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1. a) : As A is the set of all persons who are obese, while B is the set of all persons who overweight but not obese, A and B have no common elements. ...

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  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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