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Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia - AQA - A-Level Psychology - Question 5 - 2017 - Paper 1

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Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia. Th... show full transcript

Worked Solution & Example Answer:Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia - AQA - A-Level Psychology - Question 5 - 2017 - Paper 1

Step 1

Calculate the range of scores Louise gathered in both conditions of her study and complete Table 2 above.

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Answer

To calculate the range for the numbers recalled by children with dyslexia, we first identify the maximum and minimum values.

  • Maximum recalled: 17
  • Minimum recalled: 5

The range is calculated as:

Range=MaximumMinimum=175=12\text{Range} = \text{Maximum} - \text{Minimum} = 17 - 5 = 12

For children without dyslexia:

  • Maximum recalled: 17
  • Minimum recalled: 4

The range is:

Range=174=13\text{Range} = 17 - 4 = 13

Thus, the completed table should display:

  • For dyslexia: Range = 12
  • For non-dyslexia: Range = 13.

Step 2

Calculate the standard deviation for the number of pictures recalled by children with dyslexia. Show your working and give your answer to two decimal places.

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Answer

To find the standard deviation, we use the formula:

σ=(xiμ)2n\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}

Where:

  • σ\sigma = standard deviation
  • xix_i = each value
  • μ\mu = mean of the values
  • nn = number of values
  1. First, calculate the mean (μ\mu) of the scores recalled by children with dyslexia:
  • Scores: 16, 8, 5, 12, 14, 17, 11, 11, 13
  • Total = 16 + 8 + 5 + 12 + 14 + 17 + 11 + 11 + 13 = 107
  • Mean = μ=107911.89\mu = \frac{107}{9} \approx 11.89
  1. Next, calculate the squared differences from the mean:
  • For 16: (1611.89)2=17.47(16 - 11.89)^2 = 17.47
  • For 8: (811.89)2=15.29(8 - 11.89)^2 = 15.29
  • For 5: (511.89)2=47.72(5 - 11.89)^2 = 47.72
  • For 12: (1211.89)2=0.0121(12 - 11.89)^2 = 0.0121
  • For 14: (1411.89)2=4.46(14 - 11.89)^2 = 4.46
  • For 17: (1711.89)2=26.57(17 - 11.89)^2 = 26.57
  • For 11: (1111.89)2=0.7921(11 - 11.89)^2 = 0.7921
  • For 11: (1111.89)2=0.7921(11 - 11.89)^2 = 0.7921
  • For 13: $(13 - 11.89)^2 = 1.24
  1. Sum of squared differences = 17.47 + 15.29 + 47.72 + 0.0121 + 4.46 + 26.57 + 0.7921 + 0.7921 + 1.24 = 113.6221

  2. Divide by n (number of scores = 9):

113.6221912.69\frac{113.6221}{9} \approx 12.69

  1. Finally, take the square root:

σ12.693.56\sigma \approx \sqrt{12.69} \approx 3.56

Thus, the standard deviation is approximately 3.56.

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