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

The mean birth weight can be estimated using the formula:

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

Using the provided totals, we have:

$\text{Mean} = \frac{4841}{1500} \approx 3.2273$

Thus, the estimated mean birth weight is approximately ( 3.23 \text{ kg} ).

The standard deviation can be estimated using the formula:

$\text{Standard Deviation} = \sqrt{\frac{\sum f x^2}{\sum f} - (\text{Mean})^2}$

Calculating this gives:

First, we find ( \frac{15889.5}{1500} \approx 10.5923 ) and then ( (3.2273)^2 \approx 10.4156 ).

Thus,

$\text{Standard Deviation} = \sqrt{10.5923 - 10.4156} \approx \sqrt{0.1767} \approx 0.42$

Therefore, the estimated standard deviation of the birth weight is approximately ( 0.42 \text{ kg} ).

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

1. Calculate cumulative frequencies for each interval:
   • 0.0 - 1.0: 1
   • 1.0 - 2.0: 7
   • 2.0 - 2.5: 67
   • 2.5 - 3.0: 347
   • 3.0 - 3.5: 1167
   • 3.5 - 4.0: 1487
   • 4.0 - 5.0: 1497
   • 5.0 - 6.0: 1500

The median lies at the ( \frac{1500}{2} = 750 )-th value, which falls in the interval 3.0 - 3.5.

Using interpolation:

$\text{Median} = L + \frac{\frac{N}{2} - cf}{f} \times c$ Where:

• ( L = 3.0 )
• ( N = 1500 )
• ( cf = 347 )
• ( f = 820 )
• ( c = 0.5 )

Thus,

$\text{Median} = 3.0 + \frac{750 - 347}{820} \times 0.5 \approx 3.0 + 0.25 \approx 3.25$

The skewness of the distribution can be identified by comparing the mean, median, and mode:

• Since the mean (3.23) is less than the median (approximately 3.25), this indicates that the distribution is negatively skewed.

Therefore, the skewness of the distribution is negative, which is typically observed in datasets with more values concentrated on the higher end.