
David Gwyn J. answered 10/24/20
Highly Experienced Tutor (Oxbridge graduate and former tech CEO)
If the mean is 756g then the heaviest 50% weigh more than 756g.
If we go 1 standard deviation heavier (756 + 14 = 770g) the normal distribution includes a further 34.1%, or 84.1%. This means the heaviest 15.9% of fruit weight more than 770g.
We need to find the place, slightly less than 1 standard deviation, where there is 20% of the curve left. A guesstimate would therefore be 765g.
The exact number of standard deviations is a z-score, and we need to look it up in a z-score table.
The z-score table gives the area under the normal curve to the left of z, so we need 20% or 0.2 (curve is symmetrical so either side is fine). This z score is 0.845.
The equation we need is weight = mean weight + number of SDs x standard deviation weight
The weight of the fruit is 756 + 0.845 x 14 = 767.83 or 768g.
Hence the heaviest 20% of fruit weights more than (or equal to) 768g.