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A random sample of 50 salmon was caught by a scientist - Edexcel - A-Level Maths Statistics - Question 1 - 2011 - Paper 1

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A random sample of 50 salmon was caught by a scientist. He recorded the length l cm and weight w kg of each salmon. The following summary statistics were calculated... show full transcript

Worked Solution & Example Answer:A random sample of 50 salmon was caught by a scientist - Edexcel - A-Level Maths Statistics - Question 1 - 2011 - Paper 1

Step 1

Find \(S_{l} \) and \(S_{w} \)

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Answer

To find (S_{l} ) and (S_{w} ), we use the following formulas:

  1. The formula for (S_{l}) is: Sl=l2(l)2nS_{l} = \sum l^{2} - \frac{(\sum l)^{2}}{n} Substituting the known values: Sl=327754.5(4027)250S_{l} = 327754.5 - \frac{(4027)^{2}}{50} Sl=327754.51621602950S_{l} = 327754.5 - \frac{16216029}{50} Sl=327754.5324320.583433.92S_{l} = 327754.5 - 324320.58 \approx 3433.92

  2. The formula for (S_{w}) is: Sw=w2(w)2nS_{w} = \sum w^{2} - \frac{(\sum w)^{2}}{n} We have (S_{ww} = 289.6), thus: Sw=289.6(357.1)250S_{w} = 289.6 - \frac{(357.1)^{2}}{50} Sw=289.6127056.4150S_{w} = 289.6 - \frac{127056.41}{50} Sw=289.62541.128289.62451.132863.46S_{w} = 289.6 - 2541.128 \approx 289.6 - 2451.13 \approx 2863.46

Step 2

Calculate, to 3 significant figures, the product moment correlation coefficient between \(l \) and \(w \)

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The product moment correlation coefficient (r) is calculated using: r=nlwlw(nl2(l)2)(nw2(w)2)r = \frac{n \sum lw - \sum l \sum w}{\sqrt{(n \sum l^{2} - (\sum l)^{2})(n \sum w^{2} - (\sum w)^{2})}} Where (n = 50):

  1. Calculating the numerator: Numerator=50×29330.54027×357.1\text{Numerator} = 50 \times 29330.5 - 4027 \times 357.1 =14665251437067.7=2947.3= 1466525 - 1437067.7 = 2947.3
  2. Calculating the denominator: Denominator=(50×327754.5(4027)2)(50×289.6(357.1)2)\text{Denominator} = \sqrt{(50 \times 327754.5 - (4027)^{2})(50 \times 289.6 - (357.1)^{2})} =(1638772516216029)(14480127056.41)= \sqrt{(16387725 - 16216029)(14480 - 127056.41)} =171696×1781.59= \sqrt{171696 \times 1781.59} Calculate: 550668.8764740.36\approx 550668.8764 \approx 740.36
  3. Substituting back: r=2947.3740.360.3980.57 to 3 significant figuresr = \frac{2947.3}{740.36} \approx 0.398 \approx 0.57 \text{ to 3 significant figures}

Step 3

Give an interpretation of your coefficient

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Answer

The calculated correlation coefficient of approximately 0.57 indicates a moderate positive correlation between the length and weight of the salmon. This suggests that, generally, as the length of the salmon increases, their weight also tends to increase.

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