I asked 312 people how many times they visited their preferred local supermarket in a month
The results are as follows:
In the absence of the actual number of visits (I only have percentage of patrons as above), how do you calculate the mean and standard deviation for reporting purposes.
You need to be creative, because these data are consistent with any mean exceeding $0\times .05 + 1\times .07 + \cdots + 5\times .18$ = $2.89$ and any standard deviation exceeding $1.38$ (which are attained by assuming nobody visited any more than five times per month).
For reporting purposes, simply tabulate or graph the raw data:
If you must have a summary of location and spread, use alternative measures that can uniquely be found from these data. The median is between 2 and 3, because 45% visited 2 times or fewer and 67% visited 3 times or fewer. You might simply interpolate linearly and report a median of 2.3 visits per month. For the spread, use (say) an interquartile range, also computed with linear interpolation. I find Q1 is 1.4 and Q3 is 3.3, for an IQR of 1.9.
To go beyond that, you need to fit the data with a distribution, which requires assumptions and therefore is not just reporting. But it can be useful. However, these data are elusive: they will not fit standard models like Binomial or Poisson. (I recommend against trying to fit discretized versions of continuous distributions, such as Lognormal, because it's hard to find any reason why they should fit: they don't form informative bases for comparison. Moreover, since there are only six values here, it would be almost worthless to use more than one parameter in the modeling: two or more parameters give too much flexibility.)
As an example of the insight that might be afforded by a simple distributional fit, suppose the visits are made randomly over time by individuals and each individual has the same probability (per unit time) of visiting. This is potentially a useful and interesting framework against which these data can be compared. It leads to a Poisson distribution. The best fit (in a chi-squared sense) is achieved with an intensity of 3.185 per month; this also is the variance (whence the standard deviation is $\sqrt{3.185}$ = $1.8$).
This is not a good fit (as a chi-squared test will show, but the eye plainly sees): there are too many people reporting 2 visits and too few reporting 1 visit. That perhaps is the most interesting thing about this analysis. You could announce these results like this:
The median number of monthly visits among the respondents is 2.3 (with an IQR of 1.9). The data depart significantly from a (best fit) Poisson distribution with a mean of 3.18 visits per month in that 19 fewer people than expected report one visit and 37 more people than expected report two visits.
Incidentally, a Poisson fit suggestively fills in the upper tail of "5 or more visits," providing quantitative hypotheses that could be tested in follow-on surveys:
Other distributions would give different extrapolations into this upper range.
You definitely have to associate a numerical value to the class "visited five and more times a month".
By the way, I would calculate the mean and the standard deviation in the usual way. In fact, $x_i$ are your values and $p_i$ are their empirical frequency estimated on the sample. In your case
$$x_0=0 \ x_1=1 \ x_2=2 \ x_3=3 \ x_4=4 \ x_5=6$$
(you should decide $x_5$)
$$p_0=0.05 \ p_1=0.07 \ p_2=0.33 \ p_3=0.22 \ p_4=0.15 \ p_5=0.18 $$
Thus
$$\bar{x} = \sum_{i=0}^{5}x_i p_i$$
and
$$\sigma=\sqrt{\sum_{i=0}^{5}(x_i - \bar{x})^2 p_i}$$
It could be interesting to delete $x_0$ and $p_0$ and rescale all $p_i$ in order their sum is 1. So you can calculate the average number of visits to the supermarket for a person that visits the supermarket.
