The lifetime (in days) of a certain electronic component that operates in a high-temperature environment is log normally distributed with μ = 1.2 and σ = 0.4.

a. Find the mean lifetime.

b. Find the probability that a component lasts between three and six days.

c. Find the median lifetime.

d. Find the 90th percentile of the lifetimes.

Solution 1E

Step1 of 5:

Let us consider a random variable X it presents lifetime of an certain electronic components. And X follows lognormal distribution with mean and standard deviation .

That is XN(1.2, 0.42)

Here our goal is:

a).We need to find mean lifetime.

b).We need to find the probability that a component lasts between three and six days.

c).We need to find the median lifetime.

d).We need to find the 90th percentile of the lifetimes.

Step2 of 5:

a).

The probability density function of lognormal distribution is given by

Where,

mean

standard deviation

Z = standard normal variable.

Now, the mean lifetime is given by

=

=

=

= 3.5966

Hence, E(X) = 3.5966

Therefore, mean lifetime of an electric components is 3.5966.

Step3 of 5:

b).

The probability that a component lasts between three and six days is given by

=

=

=

=

=

Here is obtained from statistical table(area under normal curve)

= 0.9292 (In a standard normal table we have to see in row 1.4 under column 0.07)

= 0.4013 (In a standard normal table we have to see in row -0.2 under column 0.05)

= 0.9292 - 0.4013

= 0.5279

Hence, P(3 < X < 6) = 0.5279.

Step4 of 5:

c).

Here we need to find median lifetime of an electric component. Let us consider “m” be the median lifetime of an electric component. And we know that median divides the data into two parts that is 0.5.

Now, median of a lifetime of an electric component is given by

We know that X N(1.2, 0.42)

Take

Hence, m = 3.320.

Step5 of 5:

d).

Let us consider is a 90th percentile of lifetime of an electronic components.

That is P(X) = 0.90

Now,

Here

= 1.28

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.90 value falls, it falls in row 1.2 under column 0.08.

Hence,

90th percentile of lifetime of an electronic components is given by

=

= 1.712

= 5.5400

Therefore, 90th percentile of lifetime of an electronic components is 5.5400.