mean:
μ = ∑ x i f i / N = 4.07 \mu=\sum x_if_i/N=4.07 μ = ∑ x i f i / N = 4.07
p = μ / n = 4.07 / 7 = 0.58 p=\mu/n=4.07/7=0.58 p = μ / n = 4.07/7 = 0.58
q = 1 − p = 0.42 q=1-p=0.42 q = 1 − p = 0.42
N = 259 N=259 N = 259
n = 7 n=7 n = 7
f ( x = k ) = N C n k p k q n − k f(x=k)=NC^k_np^kq^{n-k} f ( x = k ) = N C n k p k q n − k
f ( x = 0 ) = 259 q 7 = 0.6 ≈ 1 f(x=0)=259q^{7}=0.6\approx1 f ( x = 0 ) = 259 q 7 = 0.6 ≈ 1
f ( x = 1 ) = 259 ⋅ 7 p q 6 = 5.77 ≈ 6 f(x=1)=259\cdot7pq^{6}=5.77\approx6 f ( x = 1 ) = 259 ⋅ 7 p q 6 = 5.77 ≈ 6
f ( x = 2 ) = 259 ⋅ C 7 2 p 2 q 5 = 23.91 ≈ 24 f(x=2)=259\cdot C^2_7p^2q^{5}=23.91\approx24 f ( x = 2 ) = 259 ⋅ C 7 2 p 2 q 5 = 23.91 ≈ 24
f ( x = 3 ) = 259 ⋅ C 7 3 p 3 q 4 = 55.03 ≈ 55 f(x=3)=259\cdot C^3_7p^3q^{4}=55.03\approx55 f ( x = 3 ) = 259 ⋅ C 7 3 p 3 q 4 = 55.03 ≈ 55
f ( x = 4 ) = 259 ⋅ C 7 4 p 4 q 3 = 76 f(x=4)=259\cdot C^4_7p^4q^{3}=76 f ( x = 4 ) = 259 ⋅ C 7 4 p 4 q 3 = 76
f ( x = 5 ) = 259 ⋅ C 7 5 p 5 q 2 = 62.97 ≈ 63 f(x=5)=259\cdot C^5_7p^5q^{2}=62.97\approx63 f ( x = 5 ) = 259 ⋅ C 7 5 p 5 q 2 = 62.97 ≈ 63
f ( x = 6 ) = 259 ⋅ 7 p 6 q = 28.99 ≈ 29 f(x=6)=259\cdot7p^6q=28.99\approx29 f ( x = 6 ) = 259 ⋅ 7 p 6 q = 28.99 ≈ 29
f ( x = 7 ) = 259 p 7 = 5.71 ≈ 6 f(x=7)=259p^7=5.71\approx6 f ( x = 7 ) = 259 p 7 = 5.71 ≈ 6
for fitted data:
mean:
μ = ∑ x i f i / N = 4.05 \mu=\sum x_if_i/N=4.05 μ = ∑ x i f i / N = 4.05
standard deviation:
σ = ∑ f i x i − ( ∑ f i x i ) 2 / N N = 1.32 \sigma=\sqrt{\frac{\sum f_ix_i-(\sum f_i x_i)^2/N}{N}}=1.32 σ = N ∑ f i x i − ( ∑ f i x i ) 2 / N = 1.32
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