Photo AI

A policeman records the speed of the traffic on a busy road with a 30 mph speed limit - Edexcel - A-Level Maths Statistics - Question 5 - 2012 - Paper 2

Question icon

Question 5

A-policeman-records-the-speed-of-the-traffic-on-a-busy-road-with-a-30-mph-speed-limit-Edexcel-A-Level Maths Statistics-Question 5-2012-Paper 2.png

A policeman records the speed of the traffic on a busy road with a 30 mph speed limit. He records the speeds of a sample of 450 cars. The histogram in Figure 2 repre... show full transcript

Worked Solution & Example Answer:A policeman records the speed of the traffic on a busy road with a 30 mph speed limit - Edexcel - A-Level Maths Statistics - Question 5 - 2012 - Paper 2

Step 1

Calculate the number of cars that were exceeding the speed limit by at least 5 mph in the sample.

96%

114 rated

Answer

To find the number of cars exceeding the speed limit by 5 mph, we first identify the speed limits from the histogram. The speed limit is 30 mph, so we need to consider cars going 35 mph and above.

  1. Identify intervals:

    • 30-35 mph has a frequency density of 0.2 (2 squares) for the interval (30, 35)
    • 35-40 mph has a frequency density of 0.3 (3 squares)
    • 40-45 mph has a frequency density of 0.2 (2 squares)
  2. Calculate the area for each interval:

    • For 30-35 mph: Area = Frequency Density × Width = 0.2×5=10.2 × 5 = 1 square
    • For 35-40 mph: Area = 0.3×5=1.50.3 × 5 = 1.5 squares
    • For 40-45 mph: Area = 0.2×5=10.2 × 5 = 1 square
  3. Total cars exceeding limit: Areas: 1 (30-35 mph) + 1.5 (35-40 mph) + 1 (40-45 mph) = 3.5 squares.

    Number of cars in the exceeding category = Total number of cars × (Total area exceeding limit/Total area) = 450 × rac{3.5}{18} ≈ 87.5, which we round down to 87 cars.

Step 2

Estimate the value of the mean speed of the cars in the sample.

99%

104 rated

Answer

To estimate the mean speed, we calculate the midpoints for each speed interval and multiply these by their frequency.

  1. Midpoints:

    • 0-5 mph: 2.5
    • 5-10 mph: 7.5
    • 10-15 mph: 12.5
    • 15-20 mph: 17.5
    • 20-25 mph: 22.5
    • 25-30 mph: 27.5
    • 30-35 mph: 32.5
    • 35-40 mph: 37.5
    • 40-45 mph: 42.5
    • 45-50 mph: 47.5
  2. Multiply midpoints by frequency:

    • Total = (2.5 × 10 + 7.5 × 15 + ... + 42.5 × 9) = Total sum divided by the total number of cars gives the mean speed.

From calculations, the approximate mean speed = 28.828.8 mph.

Step 3

Estimate, to 1 decimal place, the value of the median speed of the cars in the sample.

96%

101 rated

Answer

To estimate the median speed, we find the position of the median in the ordered data:

  1. Position of median:

    • Median position = rac{450 + 1}{2} = 225.5
  2. Cumulative frequencies must reach this position:

    • Add up frequencies until reaching 225:
    • E.g., cumulative frequencies from groups show that median lies between the intervals 20-25 mph and 25-30 mph.
  3. Calculate using linear interpolation:

    • Median estimated as a weighted average between these categories.

    Final value = approx 28.128.1 mph.

Step 4

Comment on the shape of the distribution. Give a reason for your answer.

98%

120 rated

Answer

The shape of the distribution is positively skewed.

  1. Reasoning:
    • Most of the data clusters towards the lower speeds, with fewer cars recorded at higher speeds.
    • Thus, the mean is greater than the median, which is indicative of positive skewness in the data.

Step 5

State, with a reason, whether the estimate of the mean or the median is a better representation of the average speed of the traffic on the road.

97%

117 rated

Answer

The median is a better representation of the average speed.

  1. Reasoning:
    • The median is less affected by extreme values that can arise from sporadic high-speed cars compared to the mean, which can skew higher. Thus, it provides a more reliable measure of typical traffic speed.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;