Question

Lifetimes of a certain brand of lightbulbs is known to follow a right-skewed distribution with mean 24 months and standard deviation 2 months. Suppose that a sample of 52 lightbulbs is taken. What is the probability that the average lifetime of these bulbs is greater than 24.32 months? Select one: a. 0.0000 b. 0.7498 c. 0.8749 d. 0.1251 e. Not enough information has been given to answer the question. f. 0.2502

Select one:

a. 0.0000

b. 0.7498

c. 0.8749

d. 0.1251

e. Not enough information has been given to answer the question.

f. 0.2502

Answer 1

Solution :

Given that,

mean = = 24 months

standard deviation = = 2 onths

n = 52

= = 24 months

= / n = 2 / 52 = 0.2773

P( > 24.32) = 1 - P( < 24.32)

= 1 - P[( - ) / < (24.32 - 24) / 0.2773]

= 1 - P(z < 1.15)

Using z table,    

= 1 - 0.8749

= 0.1251

correct option is = d