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Each member of a group of 27 people was timed when completing a puzzle - Edexcel - A-Level Maths Statistics - Question 3 - 2020 - Paper 1

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Each member of a group of 27 people was timed when completing a puzzle. The time taken, x minutes, for each member of the group was recorded. These times are summa... show full transcript

Worked Solution & Example Answer:Each member of a group of 27 people was timed when completing a puzzle - Edexcel - A-Level Maths Statistics - Question 3 - 2020 - Paper 1

Step 1

Find the range of the times.

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Answer

To find the range, we subtract the minimum time from the maximum time. From the box and whisker plot:

  • Maximum time = 68 minutes
  • Minimum time = 7 minutes

Thus, the range is calculated as follows:

Range = Maximum - Minimum = 68 - 7 = 61 minutes.

Step 2

Find the interquartile range of the times.

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Answer

The interquartile range (IQR) is determined by subtracting the first quartile (Q1) from the third quartile (Q3). From the box plot:

  • Q1 = 14 minutes
  • Q3 = 25 minutes

Therefore, IQR = Q3 - Q1 = 25 - 14 = 11 minutes.

Step 3

calculate the mean time taken to complete the puzzle.

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The mean time can be calculated using the formula:

μ=xn\mu = \frac{\sum x}{n} where ( n ) is the number of observations.

Given that x=607.5\sum x = 607.5 and there are 27 people:

μ=607.527=22.5 minutes\mu = \frac{607.5}{27} = 22.5 \text{ minutes}.

Step 4

calculate the standard deviation of the times taken to complete the puzzle.

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The standard deviation (σ) can be calculated using:

σ=x2(x)2nn\sigma = \sqrt{\frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n}}

Substituting the known values:

σ=17623.25(607.5)22727\sigma = \sqrt{\frac{17623.25 - \frac{(607.5)^2}{27}}{27}} Calculating this gives: σ=17623.2514061.2527=35562712.1 minutes.\sigma = \sqrt{\frac{17623.25 - 14061.25}{27}} = \sqrt{\frac{3556}{27}} \approx 12.1 \text{ minutes}.

Step 5

State how many outliers Taruni would say there are in these data, giving a reason for your answer.

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Answer

First calculate the mean (μ) and standard deviation (σ):

  • Mean (μ) = 22.5
  • Standard deviation (σ) = 12.1

An outlier is defined as a data point that is greater than: μ+3σ\mu + 3\sigma

Calculating: 22.5+3(12.1)=22.5+36.3=58.822.5 + 3(12.1) = 22.5 + 36.3 = 58.8

Any time greater than 58.8 is considered an outlier. From the data, only one time (the maximum time = 68 minutes) exceeds this value. Therefore, Taruni would say there is 1 outlier.

Step 6

Suggest a possible value for a and a possible value for b, explaining how your values satisfy the above conditions.

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To maintain the scenario where the median time increases but the mean remains unchanged, we select:

  • a = 45 (greater than b)
  • b = 40 (less than 45)

In this case:

  • Adding b=40 reduces the median slightly.
  • Adding a=45 keeps the average (mean) the same or slightly increases it while keeping the median in check, thus satisfying both conditions.

Step 7

Without carrying out any further calculations, explain why the standard deviation of all 29 times will be lower than your answer to part (d).

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Answer

When adding two additional times (from Adam and Beth), if these values (40 and 45) are both less than 1 standard deviation away from the mean (22.5), they are effectively bringing the overall distribution closer together. Since these values are closer to the mean than many of the existing data points, the variability of the overall dataset is reduced, leading to a lower standard deviation.

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