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The birth weights, in kg, of 1500 babies are summarised in the table below - Edexcel - A-Level Maths Statistics - Question 3 - 2010 - Paper 1

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The birth weights, in kg, of 1500 babies are summarised in the table below. Weight (kg) | Midpoint, x kg | Frequency, f 0.0 - 1.0 | 0.50 | 1 1.0 - 2.0 | 1.50 | 6 2.... show full transcript

Worked Solution & Example Answer:The birth weights, in kg, of 1500 babies are summarised in the table below - Edexcel - A-Level Maths Statistics - Question 3 - 2010 - Paper 1

Step 1

Write down the missing midpoints in the table above.

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Answer

For the ranges that are missing midpoints:

  • For 2.5 - 3.0 kg, the midpoint is 2.75 kg.
  • For 4.5 - 5.0 kg, the midpoint is 5.25 kg.

Step 2

Calculate an estimate of the mean birth weight.

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Answer

To find the mean birth weight, we use the formula:

Mean=fxf\text{Mean} = \frac{\sum f x}{\sum f}

Substituting the values:

Mean=48411500=3.2273...3.23 kg\text{Mean} = \frac{4841}{1500} = 3.2273... \approx 3.23 \text{ kg}

Step 3

Calculate an estimate of the standard deviation of the birth weight.

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Answer

The formula for standard deviation (SD) is:

SD=fx2f(fxf)2\text{SD} = \sqrt{\frac{\sum f x^2}{\sum f} - \left( \frac{\sum f x}{\sum f} \right)^2}

Substituting the values:

SD=15889.51500(3.2273...)2\text{SD} = \sqrt{\frac{15889.5}{1500} - (3.2273...)^2}

Calculating:

First, calculate the mean squared:

(3.2273...)210.4149(3.2273...)^2 \approx 10.4149

Now substituting back:

SD=10.592310.41490.17740.4210...0.42\text{SD} = \sqrt{10.5923 - 10.4149} \approx \sqrt{0.1774} \approx 0.4210... \approx 0.42

Step 4

Use interpolation to estimate the median birth weight.

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Answer

To find the median: Total frequency (n) = 1500, so we need to find ( n/2 = 750 ). Reviewing the cumulative frequency, we find:

  • 1 (0.0 - 1.0)
  • 7 (1.0 - 2.0)
  • 67 (2.0 - 2.5)
  • 347 (2.5 - 3.0)
  • 1167 (3.0 - 3.5)

Since 750 falls between 347 and 1167, the median is within the range of 3.0 - 3.5 kg. Using linear interpolation:

extMedian=3.00+(750347820347)×(3.53.0) ext{Median} = 3.00 + \left( \frac{750 - 347}{820 - 347} \right) \times (3.5 - 3.0)

Calculating:

=3.00+(403473)×0.53.00+0.42603.426 kg= 3.00 + \left( \frac{403}{473} \right) \times 0.5 \approx 3.00 + 0.4260 \approx 3.426 \text{ kg}

Step 5

Describe the skewness of the distribution. Give a reason for your answer.

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Answer

The skewness can be assessed by comparing the mean and median. Since the mean (3.23 kg) is slightly lower than the median (approximately 3.426 kg), the distribution is negatively skewed. This implies that there are a tail of lighter weights influencing the mean downward more than the median.

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