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A group of office workers were questioned for a health magazine and \( \frac{2}{5} \) found to take regular exercise - Edexcel - A-Level Maths Statistics - Question 2 - 2009 - Paper 1
Question 2
A group of office workers were questioned for a health magazine and \( \frac{2}{5} \) found to take regular exercise. When questioned about their eating habits \( \f... show full transcript
Worked Solution & Example Answer:A group of office workers were questioned for a health magazine and \( \frac{2}{5} \) found to take regular exercise - Edexcel - A-Level Maths Statistics - Question 2 - 2009 - Paper 1
Step 1
always eats breakfast and takes regular exercise
Answer
To find the probability that a randomly selected member of the group always eats breakfast and takes regular exercise, we can use the following formula:
[ P(E \cap B) = P(E | B) \cdot P(B) ]
From the information provided:
- (P(B) = \frac{2}{3}) (probability of eating breakfast)
- (P(E | B) = \frac{9}{25}) (probability of taking regular exercise given eating breakfast)
Now, substituting the values:
[ P(E \cap B) = \frac{9}{25} \cdot \frac{2}{3} = \frac{18}{75} = 0.24 ]
Thus, the probability that a randomly selected member always eats breakfast and takes regular exercise is (0.24).
Step 2
does not always eat breakfast and does not take regular exercise
Answer
To find the probability that a randomly selected member does not eat breakfast and does not take regular exercise, we first need to find (P(B')) and (P(E')):
- (P(B') = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3})
- Given (\frac{2}{5}) take regular exercise, we have:
- (P(E') = 1 - P(E) = 1 - \frac{2}{5} = \frac{3}{5})
The two events are not dependent. Therefore, we calculate:
[ P(E' \cap B') = P(E') \cdot P(B') = \frac{3}{5} \cdot \frac{1}{3} = \frac{3}{15} = \frac{1}{5} = 0.20 ]
Thus, the probability that a random member does not eat breakfast and does not take regular exercise is (0.20).
Step 3
Determine, giving your reason, whether or not always eating breakfast and taking regular exercise are statistically independent
Answer
To determine if eating breakfast and taking regular exercise are statistically independent, we need to check if:
[ P(E \cap B) = P(E) \cdot P(B) ]
From earlier calculations, we found that:
- (P(E \cap B) = 0.24)
- (P(E) = \frac{3}{5} = 0.60) (total probability of taking regular exercise)
- (P(B) = \frac{2}{3} = 0.67)
Calculating (P(E) \cdot P(B)):
[
P(E) \cdot P(B) = 0.60 \cdot 0.67 = 0.402
]
Since (P(E \cap B) \neq P(E) \cdot P(B)), they are not statistically independent.