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03 Feb 2021, 23:30
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Dropdown 1: 60% and 65%
Dropdown 2: 1 million less
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (03:51) correct
61%
(03:51)
wrong
based on 387
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The bar graph represents the percentages of the United States population, by age, that had health insurance in 1999 and 2008. The total US population in 1999 (in thousands) was 276,804; in 2008, the total U.S. population (in thousands) was 301,483.
Use the drop-down menus to complete each statement according to the information presented in the diagram.
A. Assuming independent probabilities in each year, the probability that a U.S. citizen who was 30 in 1999 had health insurance both in 1999 and in 2008 is between .
B. If 35-44 year olds represented 20% of the population in both 1999 and 2008, the number of 35-44 year olds who had health insurance in 1999 is approximately than the number of 35-44 year olds who had health insurance in 2008.
1.jpg [ 80.38 KiB | Viewed 10318 times ]
Use the drop-down menus to complete each statement according to the information presented in the diagram.
A. Assuming independent probabilities in each year, the probability that a U.S. citizen who was 30 in 1999 had health insurance both in 1999 and in 2008 is between .
B. If 35-44 year olds represented 20% of the population in both 1999 and 2008, the number of 35-44 year olds who had health insurance in 1999 is approximately than the number of 35-44 year olds who had health insurance in 2008.
Attachment:
1.jpg [ 80.38 KiB | Viewed 10318 times ]
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Official Answer
Dropdown 1: 60% and 65%
Dropdown 2: 1 million less
11 Jul 2022, 07:53
Kudos
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Hey can you provide solution
13 Jul 2022, 02:46
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dushyant099
Official Explanation
Statement 1: 60% and 65%. The probability that an individual had insurance in 1999 is equal to the percentage of the population of the same age who had insurance that year. A person who was 30 (25 to 34 years old) in 1999 (dark grey bar) therefore has a probability of approximately 78%. In 2008, this person would now be 39 years old (9 years older) and would therefore fit into the 35 to 44 year old range. Using 2008 data (light gray bar), the probability that this person has insurance in this period is approximately 80%.
Given that these two probabilities are independent, multiply to find the probability that both events occurred: 0.78 × 0.80 = 0.624 or between 60% and 65%.
Statement 2: 1 million less. To calculate the number of 35-44 year olds who had health insurance in each year, use the total population numbers given in the text accompanying the graph. First, calculate the values for each year separately. Note that the answers are given in millions (and spread relatively far apart), while the total population values are in thousands. You will need to convert but you should also be able to estimate quite a bit.
In 1999, 20% of the total population was between the ages of 35-44, and approximately 85% of individuals in that age range had health insurance (dark grey gar). Therefore, 85% of 20% of the total were 35-44 year olds with health insurance. The total of 276,804 (in thousands) is equivalent to 276,804,000 or approximately 277 million. Calculate this in one step on your calculator:
0.20 × 0.85 × 277 ≈ 47 million people between 35 and 44 had health insurance in 1999
In 2008, 20% of the total population was between the ages of 35-44, and approximately 80% of individuals in that age range had health insurance (light grey bar). This total, 301,483 (in thousands), is equivalent to approximately 301 million. Again, calculate this in one step on your calculator:
0.20 × 0.80 × 301 ≈ 48 million people between 35 and 44 had health insurance in 2008
Therefore, the number of 35-44 year olds who had health insurance in 1999 (47 million) is approximately 1 million less than the number of 35-44 year olds who had health insurance in 2008.
Statement 1: 60% and 65%. The probability that an individual had insurance in 1999 is equal to the percentage of the population of the same age who had insurance that year. A person who was 30 (25 to 34 years old) in 1999 (dark grey bar) therefore has a probability of approximately 78%. In 2008, this person would now be 39 years old (9 years older) and would therefore fit into the 35 to 44 year old range. Using 2008 data (light gray bar), the probability that this person has insurance in this period is approximately 80%.
Given that these two probabilities are independent, multiply to find the probability that both events occurred: 0.78 × 0.80 = 0.624 or between 60% and 65%.
Statement 2: 1 million less. To calculate the number of 35-44 year olds who had health insurance in each year, use the total population numbers given in the text accompanying the graph. First, calculate the values for each year separately. Note that the answers are given in millions (and spread relatively far apart), while the total population values are in thousands. You will need to convert but you should also be able to estimate quite a bit.
In 1999, 20% of the total population was between the ages of 35-44, and approximately 85% of individuals in that age range had health insurance (dark grey gar). Therefore, 85% of 20% of the total were 35-44 year olds with health insurance. The total of 276,804 (in thousands) is equivalent to 276,804,000 or approximately 277 million. Calculate this in one step on your calculator:
0.20 × 0.85 × 277 ≈ 47 million people between 35 and 44 had health insurance in 1999
In 2008, 20% of the total population was between the ages of 35-44, and approximately 80% of individuals in that age range had health insurance (light grey bar). This total, 301,483 (in thousands), is equivalent to approximately 301 million. Again, calculate this in one step on your calculator:
0.20 × 0.80 × 301 ≈ 48 million people between 35 and 44 had health insurance in 2008
Therefore, the number of 35-44 year olds who had health insurance in 1999 (47 million) is approximately 1 million less than the number of 35-44 year olds who had health insurance in 2008.
