Seeds

From a box of spruce seeds with the germination of 80%, we randomly select 10 seeds and plant them. Find the median of the random variable: the number of germinating seeds.

Correct answer:

m = 8

Step-by-step explanation:

q = 80 % = \frac{80}{100} = 0.8 r = 1 - q = 1 - 0.8 = \frac{1}{5} = 0.2 n = 10 n_{1} = (\binom{n}{1}) \cdot q^{1} \cdot (1 - q)^{n - 1} = (\binom{10}{1}) \cdot 0. 8^{1} \cdot (1 - 0.8)^{10 - 1} ≐ 4.096 \cdot 1 0^{- 6} n_{2} = (\binom{n}{2}) \cdot q^{2} \cdot (1 - q)^{n - 2} = (\binom{10}{2}) \cdot 0. 8^{2} \cdot (1 - 0.8)^{10 - 2} ≐ 0.0001 n_{3} = (\binom{n}{3}) \cdot q^{3} \cdot (1 - q)^{n - 3} = (\binom{10}{3}) \cdot 0. 8^{3} \cdot (1 - 0.8)^{10 - 3} ≐ 0.0008 n_{4} = (\binom{n}{4}) \cdot q^{4} \cdot (1 - q)^{n - 4} = (\binom{10}{4}) \cdot 0. 8^{4} \cdot (1 - 0.8)^{10 - 4} ≐ 0.0055 n_{5} = (\binom{n}{5}) \cdot q^{5} \cdot (1 - q)^{n - 5} = (\binom{10}{5}) \cdot 0. 8^{5} \cdot (1 - 0.8)^{10 - 5} ≐ 0.0264 n_{6} = (\binom{n}{6}) \cdot q^{6} \cdot (1 - q)^{n - 6} = (\binom{10}{6}) \cdot 0. 8^{6} \cdot (1 - 0.8)^{10 - 6} ≐ 0.0881 n_{7} = (\binom{n}{7}) \cdot q^{7} \cdot (1 - q)^{n - 7} = (\binom{10}{7}) \cdot 0. 8^{7} \cdot (1 - 0.8)^{10 - 7} ≐ 0.2013 n_{8} = (\binom{n}{8}) \cdot q^{8} \cdot (1 - q)^{n - 8} = (\binom{10}{8}) \cdot 0. 8^{8} \cdot (1 - 0.8)^{10 - 8} ≐ 0.302 n_{9} = (\binom{n}{9}) \cdot q^{9} \cdot (1 - q)^{n - 9} = (\binom{10}{9}) \cdot 0. 8^{9} \cdot (1 - 0.8)^{10 - 9} ≐ 0.2684 n_{10} = (\binom{n}{10}) \cdot q^{10} \cdot (1 - q)^{n - 10} = (\binom{10}{10}) \cdot 0. 8^{10} \cdot (1 - 0.8)^{10 - 10} ≐ 0.1074 n_{8} > \forall n_{i} m = 8

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