Question

1. a particular surgery for newborns has a success rate of 80%. consider the next 100 newborns who have this surgery. let x be the number of successful surgeries for the newborns. calculate the following:

a. $\mu_{x}=$
b. $\sigma_{x}=$
c. $p(x = 90)$
d. $p(x>85)$
e. $p(x < 75)$
f. $p(60\leq x\leq70)$

Explanation:

Step1: Calculate the mean ($\mu_X$)

For a binomial distribution $X\sim B(n,p)$, the mean is $\mu = np$. Here, $n = 100$ and $p=0.8$.
$$\mu_X=np=100\times0.8 = 80$$

Step2: Calculate the standard deviation ($\sigma_X$)

For a binomial distribution $X\sim B(n,p)$, the standard deviation is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 100$ and $p = 0.8$:
$$\sigma_X=\sqrt{100\times0.8\times(1 - 0.8)}=\sqrt{100\times0.8\times0.2}=\sqrt{16}=4$$

Step3: Calculate $P(X = 90)$

Using the binomial probability formula $P(X=k)={n\choose k}p^{k}(1 - p)^{n - k}$, where ${n\choose k}=\frac{n!}{k!(n - k)!}$.
$${100\choose90}(0.8)^{90}(0.2)^{10}=\frac{100!}{90!10!}(0.8)^{90}(0.2)^{10}\approx0.0077$$

Step4: Calculate $P(X>85)$

First, use the normal approximation to the binomial (since $np = 80\geq5$ and $n(1 - p)=20\geq5$). $X\sim N(np,np(1 - p))$, so $X\sim N(80,16)$.
For $X>85$, we use the continuity correction. We find $P(X>85.5)$.
The $z$-score is $z=\frac{x-\mu}{\sigma}=\frac{85.5 - 80}{4}=\frac{5.5}{4}=1.375$
$P(Z>1.375)=1 - P(Z\leq1.375)$. From the standard - normal table, $P(Z\leq1.375)\approx0.9154$
$P(X>85)\approx1 - 0.9154 = 0.0846$

Step5: Calculate $P(X<75)$

Using the normal approximation with continuity correction ($P(X<74.5)$)
$z=\frac{74.5 - 80}{4}=\frac{- 5.5}{4}=-1.375$
$P(Z<-1.375)=1 - P(Z\leq1.375)\approx1 - 0.9154 = 0.0846$

Step6: Calculate $P(60\leq X\leq70)$

Using the normal approximation with continuity correction ($P(59.5\leq X\leq70.5)$)
For $x = 59.5$, $z_1=\frac{59.5 - 80}{4}=\frac{-20.5}{4}=-5.125$
For $x = 70.5$, $z_2=\frac{70.5 - 80}{4}=\frac{-9.5}{4}=-2.375$
$P(-5.125\leq Z\leq-2.375)=P(Z\leq-2.375)-P(Z\leq-5.125)$
From the standard - normal table, $P(Z\leq-2.375)\approx0.0089$ and $P(Z\leq-5.125)\approx0$
$P(60\leq X\leq70)\approx0.0089$

Answer:

a. $\mu_X = 80$
b. $\sigma_X = 4$
c. $P(X = 90)\approx0.0077$
d. $P(X>85)\approx0.0846$
e. $P(X<75)\approx0.0846$
f. $P(60\leq X\leq70)\approx0.0089$