The normal approximation to the binomial distribution:
μ = n p = 1000 ∗ 0.003 = 3. \mu=np=1000*0.003=3. μ = n p = 1000 ∗ 0.003 = 3.
σ = n p ( 1 − p ) = 1000 ∗ 0.003 ∗ ( 1 − 0.003 ) = 1.73. \sigma=\sqrt{np(1-p)}=\sqrt{1000*0.003*(1-0.003)}=1.73. σ = n p ( 1 − p ) = 1000 ∗ 0.003 ∗ ( 1 − 0.003 ) = 1.73.
a) P ( 1.5 < X < 2.5 ) = P ( 1.5 − 3 1.73 < Z < 2.5 − 3 1.73 ) = P ( − 0.87 < Z < − 0.29 ) = P(1.5<X<2.5)=P(\frac{1.5-3}{1.73}<Z<\frac{2.5-3}{1.73})=P(-0.87<Z<-0.29)= P ( 1.5 < X < 2.5 ) = P ( 1.73 1.5 − 3 < Z < 1.73 2.5 − 3 ) = P ( − 0.87 < Z < − 0.29 ) =
= P ( Z < − 0.29 ) − P ( Z < − 0.87 ) = P ( Z < 0.87 ) − P ( Z < 0.29 ) = 0.1937. =P(Z<-0.29)-P(Z<-0.87)=P(Z<0.87)-P(Z<0.29)=0.1937. = P ( Z < − 0.29 ) − P ( Z < − 0.87 ) = P ( Z < 0.87 ) − P ( Z < 0.29 ) = 0.1937.
b) P ( X < 1.5 ) = P ( Z < 1.5 − 3 1.73 ) = P ( Z < − 0.29 ) = 0.3859. P(X<1.5)=P(Z<\frac{1.5-3}{1.73})=P(Z<-0.29)=0.3859. P ( X < 1.5 ) = P ( Z < 1.73 1.5 − 3 ) = P ( Z < − 0.29 ) = 0.3859.
c) P ( X > 2.5 ) = P ( Z > 2.5 − 3 1.73 ) = P ( Z > − 0.87 ) = 0.8078. P(X>2.5)=P(Z>\frac{2.5-3}{1.73})=P(Z>-0.87)=0.8078. P ( X > 2.5 ) = P ( Z > 1.73 2.5 − 3 ) = P ( Z > − 0.87 ) = 0.8078.
d) P ( X > 0.5 ) = P ( Z > 0.5 − 3 1.73 ) = P ( Z > − 1.45 ) = 0.9265. P(X>0.5)=P(Z>\frac{0.5-3}{1.73})=P(Z>-1.45)=0.9265. P ( X > 0.5 ) = P ( Z > 1.73 0.5 − 3 ) = P ( Z > − 1.45 ) = 0.9265.
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