The graph shows the distribution of lifespan (in years) for individuals from species 1 and species 2

If µ and σ represent mean and standard deviation of the lifespan, respectively, then, which of the following statements is true?

Correct option is D

Observations from the Graph:

Step 1: Mean (μ\muμ)

Species 1:

Using the same values as before:

• Lifespan (xxx) = $1,2,3,4$
• Frequencies (fff) = $2,8,8,2$

Mean (μ1\mu_1μ1​):

The mean is calculated using the formula:

​$$\mu = \frac{\sum (x \cdot f)}{\sum f}$$​​

For Species 1:

​$$\mu_1 = \frac{(1 \cdot 2) + (2 \cdot 8) + (3 \cdot 8) + (4 \cdot 2)}{2 + 8 + 8 + 2}\\\ \\\mu_1 = \frac{50}{20} = 2.5$$​​

Species 2:

Updated frequencies for $f=5,5,5,5$:

​$$\mu_2 = \frac{(1 \cdot 5) + (2 \cdot 5) + (3 \cdot 5) + (4 \cdot 5)}{5 + 5 + 5 + 5}\\\ \\\mu_2 = \frac{50}{20} = 2.5$$​​

Thus, the mean lifespan for both species is the same:

$$\mu_1 = \mu_2 = 2.5$$​​

Step 2: Standard Deviation (σ\sigmaσ)

The standard deviation formula is:

​$$\sigma = \sqrt{\frac{\sum f \cdot (x - \mu)^2}{\sum f}}$$​​

For Species 1:

​$$\sigma_1 = \sqrt{\frac{(2 \cdot (1 - 2.5)^2) + (8 \cdot (2 - 2.5)^2) + (8 \cdot (3 - 2.5)^2) + (2 \cdot (4 - 2.5)^2)}{20}}\\\ \\ \sigma_1 = \sqrt{\frac{13}{20}} = \sqrt{0.65} \approx 0.81$$​​

For Species 2:

​$$\sigma_2 = \sqrt{\frac{(5 \cdot (1 - 2.5)^2) + (5 \cdot (2 - 2.5)^2) + (5 \cdot (3 - 2.5)^2) + (5 \cdot (4 - 2.5)^2)}{20}}\\\ \\\sigma_2 = \sqrt{\frac{25}{20}} = \sqrt{1.25} \approx 1.12$$​​

Thus, the standard deviation for Species 2 is greater than for Species 1:

​$$\sigma_1 < \sigma_2$$​​

Final Answer:

The correct option is (d): $$\mu_1 = \mu_2 ; \sigma_1 < \sigma_2$$​​