How do I calculate the standard deviation when I have the mean and the N number

mean=7.47
n=88

You can't. You either need the deviations from the mean for each measurement or something similar from which you can calculate the deviation from the mean.

If, in your case, all 88 measurements were exactly 7.47, the standard deviation would be zero. If, however, 44 of the measurements were 6.47 and 44 were 8.47, the standard deviation would be ±1 from the following equation:

The standard deviation is calculated using

\Large \sigma = \sqrt {\frac {1}{N} \displaystyle\sum_{i=1}^N (x_i-\mu)^2}

where μ is the mean*, N is the number of samples (often you'll see 1/(N-1) instead of 1/N). The x's are the individual measurements so you can see that (x-μ) is the deviation from the mean.

*(μ is the true mean. Often you'll see \overline{x} which means the calculated mean. It's just a fine point)