SOLUTION: Pregnant women (within month 4) who were being followed-up by a nutritionist had weights (kg) equal to: 59, 66, 69, 65, 70, 55, 59, 82, 61, 74. (a) Calculate the: (i) mean? (4 m

Question 1165668: Pregnant women (within month 4) who were being followed-up by a nutritionist had
weights (kg) equal to: 59, 66, 69, 65, 70, 55, 59, 82, 61, 74.
(a) Calculate the:
(i) mean? (4 marks)
(ii) median? (4 marks)
(iii) standard deviation (7 marks)
(b) Calculate the skewness of the distribution of these weights? (3 marks)
(c) Calculate the coefficient of variation
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import skew
# Data (weights in kg)
weights = np.array([59, 66, 69, 65, 70, 55, 59, 82, 61, 74])
n = len(weights)
# (a) (i) Mean
mean_weight = np.mean(weights)
# (a) (ii) Median
median_weight = np.median(weights)
# (a) (iii) Standard Deviation (Sample, using n-1 denominator)
std_dev = np.std(weights, ddof=1)
# (b) Skewness (Sample skewness using default unbiased calculation)
skewness = skew(weights, bias=False)
# (c) Coefficient of Variation (CV)
cv = (std_dev / mean_weight) * 100
print(f"Number of observations (n): {n}")
print(f"Mean (x_bar): {mean_weight}")
print(f"Median (M): {median_weight}")
print(f"Standard Deviation (s): {std_dev}")
print(f"Skewness: {skewness}")
print(f"Coefficient of Variation (CV): {cv}")
```
```text?code_stdout&code_event_index=2
Number of observations (n): 10
Mean (x_bar): 66.0
Median (M): 65.5
Standard Deviation (s): 8.096638534327413
Skewness: 0.6688274664021394
Coefficient of Variation (CV): 12.267634142920322
```
Here are the calculated descriptive statistics for the weights of the pregnant women:
The raw data is: $59, 66, 69, 65, 70, 55, 59, 82, 61, 74$.
The sample size is $n=10$.
-----
## (a) Central Tendency and Dispersion
### (i) Mean ($\bar{x}$)
The mean is the average weight, calculated as the sum of all weights divided by the number of observations.
$$\bar{x} = \frac{\sum x}{n} = \frac{660}{10}$$
$$\mathbf{\text{Mean} = 66.0 \text{ kg}}$$
### (ii) Median ($M$)
The median is the middle value of the data when sorted.
Sorted data: $55, 59, 59, 61, \mathbf{65, 66}, 69, 70, 74, 82$
Since $n$ is even (10), the median is the average of the two middle values (the 5th and 6th values):
$$M = \frac{65 + 66}{2}$$
$$\mathbf{\text{Median} = 65.5 \text{ kg}}$$
### (iii) Standard Deviation ($s$)
The standard deviation measures the spread of the data around the mean. We calculate the **sample standard deviation** (using $n-1$ in the denominator).
$$\mathbf{\text{Standard Deviation } (s) \approx 8.097 \text{ kg}}$$
-----
## (b) Skewness of the Distribution
Skewness measures the asymmetry of the probability distribution.
$$\mathbf{\text{Skewness} \approx 0.669}$$
**Interpretation:** Since the skewness value is **positive** ($0.669 > 0$), the distribution of weights is **positively (right) skewed**. This means the tail of the distribution is longer on the right side, suggesting there are a few relatively higher weights pulling the mean above the median ($66.0 > 65.5$).
-----
## (c) Coefficient of Variation
The Coefficient of Variation (CV) expresses the standard deviation as a percentage of the mean, providing a measure of relative variability.
$$\text{CV} = \frac{s}{\bar{x}} \times 100\%$$
$$\text{CV} = \frac{8.0966}{66.0} \times 100\%$$
$$\mathbf{\text{Coefficient of Variation } (\text{CV}) \approx 12.27\%}$$
**Interpretation:** The weights have a relative variability of about **$12.27\%$** with respect to the mean weight.
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