If the mean of the data: 7, 8, 9, 7, 8, 7, $$\lambda$$, 8 is 8, then the variance of this data is:

We have eight observations in the data set, namely 7, 8, 9, 7, 8, 7, $$\lambda$$, 8.

The mean is given to be 8. By definition,

$$\text{Mean} \;(\bar{x})=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}.$$

Substituting the known values,

$$8=\dfrac{7+8+9+7+8+7+\lambda+8}{8}.$$

Multiplying both sides by 8, we get

$$64=7+8+9+7+8+7+8+\lambda.$$

Adding the numerical terms step by step,

$$7+8=15,$$

$$15+9=24,$$

$$24+7=31,$$

$$31+8=39,$$

$$39+7=46,$$

$$46+8=54.$$

So the equation becomes

$$64=54+\lambda.$$

Hence,

$$\lambda=64-54=10.$$

Now that all values are known, we proceed to find the variance. For a complete data set (population), the variance $$\sigma^{2}$$ is defined as

$$\sigma^{2}=\dfrac{\sum (x-\bar{x})^{2}}{n},$$

where $$x$$ is an individual observation, $$\bar{x}$$ is the mean, and $$n$$ is the total number of observations.

Here $$\bar{x}=8$$ and $$n=8.$$ We compute $$(x-\bar{x})^{2}$$ for each value:

For 7:  $$7-8=-1 \;\Rightarrow\; (-1)^{2}=1.$$ There are three such 7’s, so contribution $$=3\times1=3.$$

For 8:  $$8-8=0 \;\Rightarrow\; 0^{2}=0.$$ There are three such 8’s, so contribution $$=3\times0=0.$$

For 9:  $$9-8=1 \;\Rightarrow\; 1^{2}=1.$$ Contribution $$=1.$$

For 10:  $$10-8=2 \;\Rightarrow\; 2^{2}=4.$$ Contribution $$=4.$$

Adding all squared deviations,

$$\sum (x-\bar{x})^{2}=3+0+1+4=8.$$

Finally, substituting into the variance formula,

$$\sigma^{2}=\dfrac{8}{8}=1.$$

Hence, the correct answer is Option D.