The problem asks what percent of the values of a normally distributed variable $x$ are less than $(M+D)$, where $M$ is the mean and $D$ is the standard deviation.

We can use the properties of the standard normal distribution curve provided in the figure. The figure shows the percentages of data within certain standard deviations from the mean:
- Between -1 and 0 standard deviations from the mean: $34 %$
- Between 0 and +1 standard deviations from the mean: $34 %$
- Between -2 and -1 standard deviations from the mean: $14 %$
- Between +1 and +2 standard deviations from the mean: $14 %$
- Less than -2 standard deviations from the mean: $2 %$
- Greater than +2 standard deviations from the mean: $2 %$

The value $(M+D)$ corresponds to one standard deviation above the mean.
In terms of $z$-scores, this is $Z=+1$.
We want to find the percentage of values less than $(M+D)$, which means the area under the curve to the left of $Z=+1$.

We can sum the percentages from the left side of the curve up to $Z=+1$ :
- Percentage of values less than -2 standard deviations: $2 %$
- Percentage of values between -2 and -1 standard deviations: $14 %$
- Percentage of values between -1 and 0 standard deviations: $34 %$
- Percentage of values between 0 and +1 standard deviations: $34 %$

Summing these percentages:

$$
\(2 %+14 %+34 %+34 %=84 %\)
$$


Alternatively, we know that the mean (M) represents the 50th percentile. The area from the mean to one standard deviation above the mean ( M to $\(\mathrm{M}+\mathrm{D}\)$ ) is $34 %$.
So, the percentage of values less than ( $M+D$ ) is the sum of the percentage of values less than the mean ( $50 %$ ) and the percentage of values between the mean and M+D ( $34 %$ ).

$$
\(50 %+34 %=84 %\)
$$


Therefore, $84 %$ of the values of $x$ are less than $(M+D)$.

The final answer is D