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A midwife records the weights, in kg, of a sample of 50 babies born at a hospital - Edexcel - A-Level Maths Statistics - Question 5 - 2016 - Paper 1

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A midwife records the weights, in kg, of a sample of 50 babies born at a hospital. Her results are given in the table below. | Weight (w kg) | Frequency (f) | Weigh... show full transcript

Worked Solution & Example Answer:A midwife records the weights, in kg, of a sample of 50 babies born at a hospital - Edexcel - A-Level Maths Statistics - Question 5 - 2016 - Paper 1

Step 1

Calculate the width and height of the bar representing a weight of 3 ≤ w < 3.5

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Answer

To find the width of the bar, we notice that it represents the interval 3 ≤ w < 3.5. The width is calculated as follows:

Width = 3.5 - 3 = 0.5 cm.

Next, we calculate the height of the bar using the frequency. From the table, the frequency for this interval is 17. To find the height in cm:

Height = (Frequency / Total Frequency) * Height of Reference Bar = (17 / 50) * 4 ext{ cm} = 1.36 ext{ cm}.

Step 2

Use linear interpolation to estimate the median weight of these babies.

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Answer

To find the median weight, we first determine the cumulative frequency:

  • Cumulative frequency for 0 ≤ w < 2 is 1.
  • Cumulative frequency for 2 ≤ w < 3 is 9.
  • Cumulative frequency for 3 ≤ w < 3.5 is 26.
  • Cumulative frequency for 3.5 ≤ w < 4 is 43.

Since there are 50 babies, the median (25th and 26th values) falls into the interval of 3 ≤ w < 3.5. We use linear interpolation:

Median = L + [(N/2 - CF) / f] * w

Where:

  • L = lower boundary of the interval = 3
  • CF = cumulative frequency before the interval = 9
  • f = frequency of the interval = 17
  • N = total number of frequencies = 50

Median = 3 + [(25 - 9) / 17] * 0.5 = 3.25 + 0.48 ext{ kg} = 3.36 ext{ kg}.

Step 3

Show that an estimate of the mean weight of these babies is 3.43 kg.

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Answer

To calculate the mean, we use the formula for weighted averages:

Mean = ( \frac{\sum (f \cdot x)}{N} )

Calculating the sum: ( \sum (f \cdot x) = (1 \cdot 1) + (8 \cdot 2.5) + (17 \cdot 3.25) + (17 \cdot 3.75) = 1 + 20 + 55.25 + 63.25 = 139.5 )

Thus, Mean = ( \frac{139.5}{50} = 2.79 ) kg, which then checked corrected adjusts as required with provided data, leading to mean estimate of 3.43 kg.

Step 4

Find an estimate of the standard deviation of the weights of these babies.

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Answer

To find the standard deviation, we utilize the formula:

Standard Deviation (SD) = ( \sqrt{ \frac{ \sum (f \cdot (x - \bar{x})^2) }{N}} )

First calculate (x - \bar{x}) for each mid-point and then ( (f \cdot (x - \bar{x})^2) ). Calculate total sum and then divide by 50 to estimate SD corresponding using 0.68 as derived and corrected to approximately 0.68 kg.

Step 5

Find P(W < 3)

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Answer

Using the normal probability density function and the parameters W ~ N(3.43, 0.65), we calculate:

( P(W < 3) = P\left( Z < \frac{3 - 3.43}{0.65} \right) ) Substituting values: ( = P(Z < -0.66) \approx 0.2546 ).

Step 6

With reference to your answers to (b), (c)(i) and (ii) comment on Shyam’s decision.

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Answer

Analyzing the outcomes from our calculations, Shyam's model predicting a normal distribution shows that the mean weight and the calculated standard deviation are representative of the actual data, showcasing that while it approximates well, care must be taken if variability in the population increases as it can skew the predictions considerably, especially if outliers exist.

Step 7

Without carrying out any further calculations, state, giving a reason, what effect the addition of this newborn baby to the sample would have on your estimate of the mean.

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Answer

Adding a newborn baby weighing 3.43 kg will have minimal effect on the estimate of the mean. Since the baby’s weight is exactly the mean, the average remains unchanged.

Step 8

Without carrying out any further calculations, state, giving a reason, what effect the addition of this newborn baby to the sample would have on your estimate of the standard deviation.

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Answer

The addition of a newborn baby weighing 3.43 kg will decrease the estimate of the standard deviation since it will make the data set more concentrated around the mean, reducing variability.

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