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The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 100 motorists were delayed by roadworks on a stretch of motorway one Monday - Edexcel - A-Level Maths Statistics - Question 2 - 2018 - Paper 1
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The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 100 motorists were delayed by roadworks... show full transcript
Worked Solution & Example Answer:The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 100 motorists were delayed by roadworks on a stretch of motorway one Monday - Edexcel - A-Level Maths Statistics - Question 2 - 2018 - Paper 1
Step 1
Calculate the width and the height of the bar representing a delay of (11–15) minutes.
Answer
To calculate the width of the bar for the delay of (11–15) minutes, we take the class interval width:
Width = 15 - 11 = 4 minutes.
In the histogram, to find the height corresponding to this class, we first calculate the frequency density since the bar represents the number of motorists.
Frequency density = ( \frac{f}{\text{width}} = \frac{12}{4} = 3 ) motorists per minute.
Thus, the height of the bar = 3 cm.
Step 2
Use linear interpolation to estimate the median delay.
Answer
To estimate the median delay, we first find the cumulative frequency. The median is at the position ( \frac{n}{2} = \frac{100}{2} = 50 ) in the cumulative frequency distribution.
Cumulative frequencies:
- For (3–6): 38
- For (7–8): 38 + 25 = 63 (median lies here) Thus, the median class is (7–8).
Using linear interpolation:
- Lower boundary of median class = 7
- Frequency of median class = 25
- Cumulative frequency before median class = 38
Median = Lower boundary + ( rac{(\frac{n}{2} - \text{cumulative frequency before})}{f} \cdot ext{class width} ) = 7 + ( \frac{(50 - 38)}{25} \cdot 1 ) = 7 + 0.48 = 7.48 (approx).
Step 3
Calculate an estimate of the mean delay.
Answer
To find the mean delay, we can use the formula:
( \text{Mean} = \frac{\sum (f \cdot x)}{n} )
Where:
- ( f ) = number of motorists
- ( x ) = delay midpoint
- ( n = 100 )
Calculating ( f \cdot x ):
- (3–6): 38 * 4.5 = 171
- (7–8): 25 * 7.5 = 187.5
- (9–10): 18 * 9.5 = 171
- (11–15): 12 * 13 = 156
- (16–20): 7 * 18 = 126
Total ( \sum (f \cdot x) = 171 + 187.5 + 171 + 156 + 126 = 811.5 )
Thus, Mean = ( \frac{811.5}{100} = 8.115 = 8.12 ) (approx).
Step 4
Calculate an estimate of the standard deviation of the delays.
Answer
The standard deviation (( , \sigma )) is calculated using:
( \sigma = \sqrt{\frac{\sum f(x - \text{mean})^2}{n}} )
Using the mean calculated earlier (8.12):
- Calculate ( (x - \text{mean})^2 ) for each class:
- (3–6): ( (4.5 - 8.12)^2 \cdot 38 )
- (7–8): ( (7.5 - 8.12)^2 \cdot 25 )
- (9–10): ( (9.5 - 8.12)^2 \cdot 18 )
- (11–15): ( (13 - 8.12)^2 \cdot 12 )
- (16–20): ( (18 - 8.12)^2 \cdot 7 )
Calculate the total sum and divide by 100, then take the square root to find ( \sigma = 3.89 ) (approx).
Step 5
Evaluate this coefficient for the above data, giving your answer to 2 significant figures.
Answer
To find the coefficient of skewness: [ \text{Skewness} = \frac{3(\text{mean} - \text{median})}{\text{standard deviation}} ] Substituting the known values: Mean = 8.12, Median = 7.48, Standard Deviation = 3.89: [ = \frac{3(8.12 - 7.48)}{3.89} \approx 0.49 \text{(to 2 significant figures)} ]
Step 6
State, giving a reason, how the delays on this stretch of motorway on Friday are different from the delays on Monday.
Answer
On Friday, the coefficient of skewness is -0.22, indicating a negative skew. This suggests that there are fewer longer delays compared to Monday's delays, which have a positive skew (0.49). In practical terms, the delays on Friday tend to be shorter and more clustered around the lower end, contrasting with the longer delays observed on Monday.