Question

2. The probability that an electronic component will faill in performance is 0.2 Use the normal approximation to Binomial to find the probability that among 100 such components, (a) at least 23 will fail in performance. (b) between 18 and 26 (inclusive) will fail in performance. That is find P(18 26) X

Answer 1

(a)

Lnp 100X 0.2 20) =

Vnpq 100 x 0.2 x 0.84 =

To find P(X$\geq$23):
Applying Continuity Correction:

P(X>22.5):

Z = (22.5 - 20)/4

= 0.625

Table gives area= 0.2357

So,

P(X$\geq$23) = 0.5 - 0.2357 = 0.2643

So,

Answer is:

0.2643

(b)

To find P(18$\leq$X$\leq$23):
Applying Continuity Correction:

To find P(17.5 < X < 23.5):

Case 1: For X from 17.5 to mid value:

Z = (17.5 - 20)/4

= - 0.625

Table gives area= 0.2357

Case 2: For X from mid value to 23.5:

Z = (23.5 - 20)/4

= 0.875

Table gives area= 0.3106

So,

P(18$\leq$X$\leq$23):= 0.2357 +0.3106 = 0.5463

So,

Answer is:

0.5463