Answer to Question #211695 in Statistics and Probability for 5678

Question

1.      (i) Use normal approximation to a Binomial Distribution (X) having n=500 trials and p = 0.025 for finding out:

(a)   P(X>10);                              (b) P(X<18)       (c) P(X>21)               (d) P(9<X<14)    (4)

        

Answer 1

Binomial distribution bin(n,p) can be estimated using a normal distribution norm(np,np(1-p). However, correction for continuity should be carried out. Below is the correction factor table.

$If \,P(X=y) \,use \,P(y-9.5<X<y+0.5)$

$If \,P(X>y) \,use \,P(X>y+0.5)$

$If \,P(X<) \,use \,P(X<y-0.5)$

$If\, P(X\le) \,use\, P(X< y+0.5)$

$If \, P(X\ge) \,use \,P(X>y-0.5)$

Before using normal approximation for binomial distribution the conditions $np\ge5$ or $n(1-p)\ge5$ should be met.

$n=500$

$P=0.025$

$np=500\times0.025=12.5$

$n(1-p)=500\times0.9725=487.5$

Both np and n(1-p) are greater that 5. Thus, we can use normal approximation.

$np=125$

$\sqrt{np(1-p)}=3.49106$

(a) $P(X>10)$

Correcting for continuity, $P(X>10)$ becomes $P(X>10.5)$ based on the above table.

$P(>10.5)=1-P(X<10.5)$

$=1-P(z<\frac{10.5-12.5}{3.49106})$

$=1-P(z<-0.573)$

$=1-0.28336$

$=0.71664$

(b). $P(X<18)$

Correcting for continuity becomes $P(X<17.5)$

$P(X<17.5)=P(z<\frac{17.5-12.5}{3.49106})$

$=P(z<1.432)$

$=0.924$

(c) $P(X>21)$

Correcting for continuity, $P(X>21)$ becomes $P(X>21.5)$

$P(>21.5)=1-P(X<21.5)$

$=1-P(z<\frac{21.5-12.5}{3.49106})$

$=1-P(z<2.578)$

$=1-0.995$

$0.005$

(d) $P(9<X<14)$

$P(9<X<14)=P(X<14)-P(X<9)$

Correcting for continuity, $P(9<X<14)$ becomes $P(X<13.5)-P(X<8.5)$

$P(X<13.5)-P(X<8.5)=P(z<\frac{13.5-12.5}{3.49106})-P(z<\frac{8.5-12.5}{3.49106})$

$=P(z<0.286)-P(z<-1.146)$

$=0.6127-0.1259$

$=0.4868$