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)   

        

    

Answer 1

$p=0.025 \\ \mu= np \\ mu= 50 \times 0.025= 12.5 \\ \sigma= \sqrt{np(1-p)} \\ \sigma= \sqrt{12.5(1-0.025)} = 3.49$

With continuity correction factor

(a)

$P(X>10) = P(X>10.5) \\ = 1 -P(X<10.5) \\ = 1 -P(Z< \frac{10.5-12.5}{3.49}) \\ = 1 -P(Z< -0.5730) \\ = 1 -0.2833 \\ =0.7167$

(b)

$P(X<18) = P(X<17.5) \\ = P(Z < \frac{17.5-12.5}{3.49}) \\ = P(Z< 1.4326) \\ = 0.9240$

(c)

$P(X>21) = P(X>21.5) \\ = 1 -P(X<21.5) \\ = 1 -P(Z< \frac{21.5-12.5}{3.49}) \\ = 1 -P(Z< 2.5787) \\ = 1 -0.9950 \\ = 0.005$

(d)

$P(9<X<14) = P(8.5<X<14.5) \\ = P(X<14.5) -P(X<8.5) \\ =P(Z< \frac{14.5-12.5}{3.49}) -P(Z< \frac{8.5-12.5}{3.49}) \\ = P(Z< 0.5730) -P(Z< -1.1461) \\ = 0.7166 -0.1258 \\ = 0.5908$