Answer :
Let's go through the options one by one to see which statement is true based on the data provided in the two-way table.
Option A:
[tex]\( P(\text{consumes } 1,000-1,500 \text{ calories} \mid \text{weight is } 165) \)[/tex]
This means the probability of consuming 1,000-1,500 calories given that the person weighs 165 lb. We compare this to the general probability of consuming 1,000-1,500 calories.
- From the table, for those who weigh 165 lb, 15 individuals consume 1,000-1,500 calories out of a total of 117 individuals who weigh 165 lb.
- So, [tex]\( P(\text{consumes } 1,000-1,500 \text{ calories} \mid \text{weight is } 165) = \frac{15}{117} \)[/tex].
The overall probability of consuming 1,000-1,500 calories is:
- Total in 1,000-1,500 range = 140
- Total individuals = 500
- So, [tex]\( P(\text{consumes } 1,000-1,500 \text{ calories}) = \frac{140}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Option B:
[tex]\( P(\text{weight is 120 lb and consumes } 2,000-2,500 \text{ calories}) \neq P(\text{weight is 120 lb}) \)[/tex]
- 10 individuals who weigh 120 lb consume 2,000-2,500 calories out of a total of 180 individuals who weigh 120 lb.
- [tex]\( P(\text{weight is 120 lb and consumes } 2,000-2,500) = \frac{10}{500} \)[/tex].
- [tex]\( P(\text{weight is 120 lb}) = \frac{180}{500} \)[/tex].
Since these probabilities are different, this statement is true.
Option C:
[tex]\( P(\text{weight is 165 lb} \mid \text{consumes } 1,000-2,000 \text{ calories}) = P(\text{weight is 165 lb}) \)[/tex]
- First, find [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) \)[/tex]:
- Total for 1,000-2,000 range (sum of 1,000-1,500 and 1,500-2,000) = 140 + 250 = 390
- Total individuals = 500
- [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) = \frac{390}{500} \)[/tex].
- For those who consume 1,000-2,000 calories and weigh 165 lb, it's:
- [tex]\( P(\text{weight is 165} \mid \text{consumes } 1,000-2,000) = \frac{15 + 27}{390} = \frac{42}{390} \)[/tex].
[tex]\( P(\text{weight is 165 lb}) = \frac{117}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Option D:
[tex]\( P(\text{weight is 145 lb} \mid \text{consumes } 1,000-2,000 \text{ calories}) = P(\text{consumes } 1,000-2,000 \text{ calories}) \)[/tex]
- For those who consume 1,000-2,000 calories and weigh 145 lb:
- [tex]\( P(\text{weight is 145 lb} \mid \text{consumes } 1,000-2,000) = \frac{35 + 143}{390} = \frac{178}{390} \)[/tex].
- As calculated earlier, [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) = \frac{390}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Therefore, the true statement is Option B.
Option A:
[tex]\( P(\text{consumes } 1,000-1,500 \text{ calories} \mid \text{weight is } 165) \)[/tex]
This means the probability of consuming 1,000-1,500 calories given that the person weighs 165 lb. We compare this to the general probability of consuming 1,000-1,500 calories.
- From the table, for those who weigh 165 lb, 15 individuals consume 1,000-1,500 calories out of a total of 117 individuals who weigh 165 lb.
- So, [tex]\( P(\text{consumes } 1,000-1,500 \text{ calories} \mid \text{weight is } 165) = \frac{15}{117} \)[/tex].
The overall probability of consuming 1,000-1,500 calories is:
- Total in 1,000-1,500 range = 140
- Total individuals = 500
- So, [tex]\( P(\text{consumes } 1,000-1,500 \text{ calories}) = \frac{140}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Option B:
[tex]\( P(\text{weight is 120 lb and consumes } 2,000-2,500 \text{ calories}) \neq P(\text{weight is 120 lb}) \)[/tex]
- 10 individuals who weigh 120 lb consume 2,000-2,500 calories out of a total of 180 individuals who weigh 120 lb.
- [tex]\( P(\text{weight is 120 lb and consumes } 2,000-2,500) = \frac{10}{500} \)[/tex].
- [tex]\( P(\text{weight is 120 lb}) = \frac{180}{500} \)[/tex].
Since these probabilities are different, this statement is true.
Option C:
[tex]\( P(\text{weight is 165 lb} \mid \text{consumes } 1,000-2,000 \text{ calories}) = P(\text{weight is 165 lb}) \)[/tex]
- First, find [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) \)[/tex]:
- Total for 1,000-2,000 range (sum of 1,000-1,500 and 1,500-2,000) = 140 + 250 = 390
- Total individuals = 500
- [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) = \frac{390}{500} \)[/tex].
- For those who consume 1,000-2,000 calories and weigh 165 lb, it's:
- [tex]\( P(\text{weight is 165} \mid \text{consumes } 1,000-2,000) = \frac{15 + 27}{390} = \frac{42}{390} \)[/tex].
[tex]\( P(\text{weight is 165 lb}) = \frac{117}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Option D:
[tex]\( P(\text{weight is 145 lb} \mid \text{consumes } 1,000-2,000 \text{ calories}) = P(\text{consumes } 1,000-2,000 \text{ calories}) \)[/tex]
- For those who consume 1,000-2,000 calories and weigh 145 lb:
- [tex]\( P(\text{weight is 145 lb} \mid \text{consumes } 1,000-2,000) = \frac{35 + 143}{390} = \frac{178}{390} \)[/tex].
- As calculated earlier, [tex]\( P(\text{consumes } 1,000-2,000 \text{ calories}) = \frac{390}{500} \)[/tex].
Since these probabilities are not equal, this statement is false.
Therefore, the true statement is Option B.
