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The marks, x, of 45 students randomly selected from those students who sat a mathematics examination are shown in the stem and leaf diagram below - Edexcel - A-Level Maths Statistics - Question 4 - 2012 - Paper 1

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The marks, x, of 45 students randomly selected from those students who sat a mathematics examination are shown in the stem and leaf diagram below. Mark Totals 3 6 9... show full transcript

Worked Solution & Example Answer:The marks, x, of 45 students randomly selected from those students who sat a mathematics examination are shown in the stem and leaf diagram below - Edexcel - A-Level Maths Statistics - Question 4 - 2012 - Paper 1

Step 1

Write down the modal mark of these students.

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Answer

To find the modal mark, we need to identify the mark that appears most frequently. From the stem and leaf diagram, we see:

  • The mark 6 appears 6 times (in the totals under stem 6), and thus it is the most frequent mark.
  • Therefore, the modal mark is 6.

Step 2

Find the values of the lower quartile, the median and the upper quartile.

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Answer

To find the quartiles, we will first list the marks in ascending order:

  • 36, 36, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 46, 46, 56, 56, 56, 58, 58, 66, 66, 66, 66, 67, 67, 69, 70, 70, 71, 73, 74, 75, 78, 79, 80, 81, 81, 81, 81, 81, 82, 82, 83, 84.

  • The total number of marks is 45.

  • Lower Quartile (Q1) is the 11th value: 46.

  • Median (Q2) is the 23rd value: 55.

  • Upper Quartile (Q3) is the 34th value: 66.

Step 3

Find the mean and the standard deviation of the marks of these students.

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Answer

The mean (μ\mu) can be calculated using the formula:

μ=xn=24974555.49\mu = \frac{\sum x}{n} = \frac{2497}{45} \approx 55.49

To find the standard deviation (sd), we use:

sd=x2n(μ)2=14336945(55.49)210.45 sd = \sqrt{\frac{\sum x^2}{n} - (\mu)^2} = \sqrt{\frac{143369}{45} - (55.49)^2} \approx 10.45

Step 4

Describe the skewness of the marks of these students, giving a reason for your answer.

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Answer

The relationship between the mean, median, and mode can be used to assess skewness:

  • Since mean<median<mode\text{mean} < \text{median} < \text{mode}, this indicates that the distribution is negatively skewed.
  • We conclude that there is a negative skewness in the data since the mean is less than the median.

Step 5

Find the mean and standard deviation of the scaled marks of all these students.

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Answer

First, we scale the marks. Scaling involves subtracting 5 from each mark and then further reducing this by 10%. The new mean can be calculated as:

New Mean=(555)×0.9=45 (after scaling)\text{New Mean} = (55 - 5) \times 0.9 = 45 \text{ (after scaling)}

For the standard deviation:

New SD=10×0.9=9\text{New SD} = 10 \times 0.9 = 9

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