Question

Suppose it is known that 20% of batteries have a lifespan shorter than the advertised lifespan. Suppose that 100 batteries are selected at random. What is the approximate probability (using the continuity correction) that at least 10 batteries will have a short lifespan?

Answer 1

Solution :

Given that,

p = 20% = 0.20

q = 1 - p = 1 - 0.20 = 0.80

n = 100

Using binomial distribution,

= n * p = 100 * 0.20 = 20

= n * p * q = 100 * 0.20 * 0.80 = 4

Using continuity correction ,

P(x 9.5) = 1 - P(x 9.5)

= 1 - P((x - ) / () / )

= 1 - P(z -2.625)

= 1 - 0.0043   

= 0.9957

Probability = 0.9957