Question

The Karine Light Company manufactures inexpensive light-emitting diode (LED) bulbs whose estimated lifetime isnormally distributed with a mean of 9 200 hours and a standard deviation of 600 hours. Assume that they are fightedcontinuously:(a) Calculate the probability that the LED will fail within one year.(b) Calculate the probability that the LED will still be operating 400 days after installation.(c) Out of 100 LEDs. about how many will have lifetimes between 8 500 and 10 200 hours?

Answer 1

a) P = 0.2327; b) P = 0.25143 c) 83 LEDs

Step-by-step explanation:

Let

μ = mean ;

σ = standard deviation;

X = lifetime(hours) ;

a)

We have relation

Z = (X-μ)/σ ~ N(0,1)

Now,

we know that there are 365 days in year

hours in one year = 365 *24 = 8760 hours

Also μ = 9200hrs, σ = 600 hrs;

The equation for probability of LED failing in 1 year is:

P ( X≤8760 ) = P[(X-μ)/σ≤(8760-μ)/σ] --------------------equation (1)

By putting values in equation 1 we get,

P ( X≤8760 ) = P[Z≤(8760-9200)/600]

P ( X≤8760 ) = P[Z≤-0.73333]

From standard normal distribution table, we can determine the value to be

P = 0.2327

Therefore the probability of LED failing within a year is 0.2327

b)

For 400 days,

hours in 400 days = 400 * 24 = 9600 hours

P ( X≤9600 ) = P[(X-μ)/σ≤(9600-μ)/σ] --------------------equation (2)

By putting values in equation 2 we get,

P ( X≤9600 ) = P[Z≤(9600-9200)/600]

P ( X≤9600 ) = P[Z>0.67] = 1-P[Z≤0.67]

From standard normal distribution table, we can determine the value to be

P = 1-0.74857 = 0.25143

Therefore the probability of LED failing within 400 days is 0.25143

c)

Now to calculate lifetime using the relation

P ( 8500≤X≤10200 ) = P[(8500-μ)/σ≤(X-μ)/σ≤(10200-μ)/σ]

P ( 8500≤X≤10200 ) = P[(8500-μ)/σ≤Z≤(10200-μ)/σ] ----------equation 93)

Now substituting values in equation 3 we get,

P ( 8500≤X≤10200 ) = P[(8500-9200)/600 ≤ Z ≤ (10200-9200)/600]

P ( 8500≤X≤10200 ) = P[-1.17 ≤ Z > 1.67 ]

P ( 8500≤X≤10200 ) = P[Z ≤ 1.67] - P[Z ≤ -1.17 ]

Now from standard normal distribution table, we get

P = 0.95252 - 0.121

P = 0.831540

So, the probability of LED's lifetime to be between 8500 and 10200 hrs in 0.831540

Now finally to calcualte the number of LEDs having this lifetime,

Number of LEDs = total LEDs * probability

= 100 * 0.831540

= 83 LEDs