Normal Distribution! Please help!

Weights, X kg, of 10 year old boys are normally distributed. Proportion of boys weighing less than 25kg is 35% and more than 35kg is also 35%. (Mean is therefore 30kg) Calculate Standard deviation of X.

So 65% weigh less than 35 kg.
From tables this means that 35 kg is 0.385 standard deviations above the mean.
So 0.385 standard deviations is 5 kg.
5 ÷ 0.385 = 12.99
Standard deviation is approximately 13 kg.

Response to additional details
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In a book of tables I found a table of the cumulative frequency Φ(x) for the Normal distribution. This shows that 50% of the distribution is 0 standard deviations below the mean (obviously) and 84.1% is less than 1 standard deviation above the mean and 65% is less than 0.385 standard deviations above the mean.

Since weight percentage is 35, we can say the following:

p = 0.35
q = 1 - 0.35 (since q = 1 - p)
q = 0.65

Mean is 30 kg.
Formula for mean is u = np
30 = n x 0.35
n = 85.71
n ~ 86

So variance = npq
variance = 86 x 0.35 x 0.65
variance = 19.565

Since standard deviation is the square root of variance, we would get the following:

Standard deviation = sq root (19.565)
Standard deviation = 4.423
________

I hope this helps!