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The table shows some information about the dress sizes of 25 women - Edexcel - GCSE Maths - Question 3 - 2017 - Paper 3
Question 3
The table shows some information about the dress sizes of 25 women. | Dress size | Number of women | |------------|-----------------| | 8 | 2 ... show full transcript
Worked Solution & Example Answer:The table shows some information about the dress sizes of 25 women - Edexcel - GCSE Maths - Question 3 - 2017 - Paper 3
Step 1
Find the median dress size.
Answer
To find the median dress size, we first need to list all the dress sizes in order, considering the frequency of each size:
- Size 8 appears 2 times.
- Size 10 appears 9 times.
- Size 12 appears 8 times.
- Size 14 appears 6 times.
This gives us the ordered list:
8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 14, 14.
Since there are 25 data points (an odd number), the median will be the value at position:
[ \text{Median Position} = \frac{25 + 1}{2} = 13 ]
Looking at the ordered list, the 13th value is 12. Thus, the median dress size is 12.
Step 2
Is Zoe correct?
Answer
To assess Zoe's statement, we need to calculate the probability that a randomly chosen woman has either a shoe size of 7 or a dress size of 14.
Given that:
- 3 women have a shoe size of 7.
- 6 women have a dress size of 14.
Since these two events are not mutually exclusive (it is possible for a woman to belong to both categories), we use the formula for the probability of the union of two events:
[ P(A \cup B) = P(A) + P(B) - P(A \cap B) ]
Let:
- (A) be the event of having a shoe size of 7.
- (B) be the event of having a dress size of 14.
We know:</n- (P(A) = \frac{3}{25})
- (P(B) = \frac{6}{25})
- We don't have information on the intersection (P(A \cap B)), but since we know Zoe stated the probability is (\frac{9}{25}) without considering overlaps correctly, we conclude she is incorrect unless more information is provided about women with both attributes. The final conclusion is that at least one of the events must overlap for her probability to be accurate.