In: Statistics and Probability
TKK Products manufactures 50-, 60-, 75-, and 100-watt electric light bulbs. Laboratory tests show that the lives of these light bulbs are normally distributed with a mean of 650 hr and standard deviation of 100 hr. What is the probability that a TKK light bulb selected at random will burn for the following times? (Round your answers to four decimal places.)
(a) more than 850 hr
(b) less than 500 hr
(c) between 650 and 850 hr
(d) between 500 and 725 hr
Solution :
Given that ,
mean = = 650
standard deviation = = 100
P(X > 850) = 1 - P(x < 850)
= 1 - P((x - ) / < (850 - 650) / 100)
= 1 - P(z < 2) Using standard normal table,
= 1 - 0.9772
= 0.0228
P(x > 850) = 0.0228
Probability = 0.0228
b)
P(x < 500) = P((x - ) / < (500 - 650) / 100) = P(z < -1.5)
P(x < 500) = 0.0668
Probability = 0.0668
c)
P(650 < x < 850) = P((650 - 650) / 100) < (x - ) / < (850 - 650) / 100) )
P(650 < x < 850) = P(0 < z < 2)
P(650 < x < 850) = P(z < 2) - P(z < 0)
P(650 < x < 850) = 0.9772 - 0.5
Probability = 0.9772 - 0.5 = 0.4772
d)
P(500 < x < 725) = P((500 - 650) / 100) < (x - ) / < (725 - 650) / 100) )
P(650 < x < 850) = P(-1.5 < z < 0.75)
P(650 < x < 850) = P(z < 0.75) - P(z < -1.5)
P(650 < x < 850) = 0.7734 - 0.0668 = 0.7066
Probability = 0.7066