Suppose that x has a binomial distribution with n = 50 and p = .6, so that μ = np = 30 and σ = np(1 − p) = 3.4641. Calculate the following probabilities using the normal approximation with the continuity correction. (Hint: 22 < x < 37 is the same as 23 ≤ x ≤ 36. Round your answers to four decimal places.) g

Question:Suppose that x has a binomial distribution with n = 50 and p = 0.6, so that μ = np = 30 and σ = np(1 − p) = 3.4641. Approximate the following probabilities using the normal approximation with the continuity correction. (Hint: 26 < x < 32 is the same as 27 ≤ x ≤ 31. Round your answers to four decimal places.)(a)    P(x = 30)(b)    P(x = 26)(c)    P(x ≤ 26)(d)    P(26 ≤ x ≤ 32)(e)    P(26 < x < 32)Answer:(a)    P(x = 30)=0.1113(b)    P(x = 26)=0.8064(c)    P(x ≤ 26)=0.9032(d)    P(26 ≤ x ≤ 32)=0.6674 (e)    P(26 < x < 32)=0.5101 Step-by-step explanation:Here , X has a binomial distribution with n= 50 and p=6The mean is μ=np=50(0.6)=30The standard deviation is [tex]\sigma=\sqrt{np(1-p)}=\sqrt{50(0.6)(0.4)[/tex]=>3.4641(A)  P(x = 30)=>[tex]P(X_{Binomial}=30)= P(30-0.5\leq{normal}\leq30+0.5)[/tex]=>[tex]P(29.5\leqY_{normal}\leq30.5)[/tex]=>[tex]P(\frac{29.5-30}{3.4641}\leq\frac{Y_{normal}-\mu}{\sigma}\leq\frac{30.5-30}{3.4641})[/tex]=>[tex]P(-0.14\leqZ\leq0.14)[/tex]=>[tex]P(z\leq0.14)- P(z\leq0.14)[/tex]=>[=NORMSDIST(0.14)]-[=NORMSDIST(0.14)]=>0.5557-0.4443=>0.1113(B)P(x = 26)=>[tex]P(X_{Binomial}=26)= P(26-0.5\leqY_{normal}\leq26+0.5)[/tex]=>[tex]P(25.5\leqY_{normal}\leq26.5)[/tex]=>[tex]P(\frac{25.5-30}{3.4641}\leq\frac{Y_{normal}-\mu}{\sigma}\leq\frac{26.5-30}{3.4641})[/tex]=>[tex]P(-1.30\leq z \leq1.30)[/tex]=>[tex]P(Z\leq1.30)- P(Z\leq1.30)[/tex]=>[=NORMSDIST(1.30)]-[=NORMSDIST(-1.30)]=>0.9032-0.0968=>0.8064  (c) P(x ≤ 26)=>[tex]P(X_{Binomial}\leq26)= P(Y_{normal}\leq26+0.5)[/tex]=>[tex]P(Y_{normal}\leq26.5)[/tex]=>[tex]P(\frac{Y_{normal}- \mu}{\sigma}\leq\frac{26.5-30}{3.4641})[/tex]=>[tex]P(Z\leq1.30)[/tex]=>[tex]P(Z\leq1.30)[/tex]=>[=NORMSDIST(1.30)]=>0.9032D)  P(26 ≤ x ≤ 32)=>[tex]P(26\leqX_{Binomial}\leq32)= P(26-0.5\leqY_{normal}\leq32+0.5)[/tex]=>[tex]P(25.5\leqY_{normal}\leq32.5)[/tex]=>[tex]P(\frac{25.5-30}{3.4641}\leq\frac{Y_{normal}-\mu}{\sigma}\leq\frac{32.5-30}{3.4641})[/tex]=>[tex]P(-1.30\leqZ\leq0.72)[/tex]=>[tex]P(Z\leq0.72)- P(Z\leq-1.30)[/tex]=[=NORMSDIST(0.72)]-[=NORMSDIST(-1.30)]=>0.7642 -0.0968=>0.6674  (E) P(26 < x < 32)=>[tex]P(26<X_{Binomial}<32)= P(26-0.5<Y_{normal}<32-0.5)[/tex]=>[tex]P(26.5<Y_{normal}<=31.5)[/tex]=>[tex]P(\frac{26.5-30}{3.4641}<\frac{Y_{normal}-\mu}{\sigma}<\frac{31.5-30}{3.4641})[/tex]=>[tex]P(-1.01\leqz\leq0.43)[/tex]=>[tex]P(Z\leq0.43)- P(Z\leq-1.01)[/tex]=>[=NORMSDIST(0.43)]-[=NORMSDIST(-1.01)]=>0.6664-0.1563=>0.5101