Question 253060
This is a binomial distribution. B(12,0.6)
1.Exactly 10 seeds is 12C10 (0.6)^10 x (0.4)^2
this equates to 66 x 0.0060466 x 0.16 = 0.0638 or 6.38%
2. All seeds is 12C12 x (0.6)^12 x 1 = 0.00217678 or 0.22%
3. 10 or more are the 10, 11 and 12 cases added together. the case of 11 is
12C11 x (0.6)^11 x 0.4 = 12 x 0.0036797 x 0.4 = 0.017414 or 1.7414 %
So the probability of 10 or more is 6.38 + 1.74 + 0.22 % = 8.34% 
4. The mean is the number of trials x probability or 12 x 0.6 = 7.2 successes
5. The variance is n x p x (1-p) which in this case is 12 x 0.6 x 0.4 = 2.88.
The standard deviation is the sqrt of this or 1.697.