Let n = 15, p = 25% = 0.25, q = 1 - p = 0.75
According to the binomial distribution:
P(x=k)=Cnkpkqn−k
a) P(3≤x≤6)=∑k=36P(x=k)==∑k=36Cnkpkqn−k=1⋅2⋅315⋅14⋅130.2530.7512+1⋅2⋅3⋅415⋅14⋅13⋅120.2540.7511++1⋅2⋅3⋅4⋅515⋅14⋅13⋅12⋅110.2550.7510++1⋅2⋅3⋅4⋅5⋅615⋅14⋅13⋅12⋅11⋅100.2560.759==0.225199065+0.225199065++0.165145981+0.091747767==0.707291878≈70.73%
b) P(x<4)=∑k=03P(x=k)==∑k=03Cnkpkqn−k=0.2500.7515+1150.2510.7514+1⋅215⋅140.2520.7513++1⋅2⋅315⋅14⋅130.2530.7512==0.013363461+0.066817305++0.155907045+0.225199065==0.461286876≈46.13%
c) P(x>5)=∑k=615Cnkpkqn−k=(p+q)n−∑k=05Cnkpkqn−k=1−0.461286876−1⋅2⋅3⋅415⋅14⋅13⋅120.2540.7511++1⋅2⋅3⋅4⋅515⋅14⋅13⋅12⋅110.2550.7510==0.538713124−0.225199065−0.165145981==0.148368078≈14.84%
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