Question

The lifetime of a light bulb is exponentially distributed with a mean life of
18
months. If there are
60
%
chances that a light bulb will last at most
t
months, then what is the value of
t
?

Answer 1

Answer:

The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.

Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.

Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts.

Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.

X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m, the decay parameter.

m=\frac{1}{\mu }. Therefore, m=\frac{1}{4}=0.25.

The standard deviation, σ, is the same as the mean. μ = σ

The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25).

The probability density function is f(x) = me–mx. The number e = 2.71828182846… It is a number that is used often in mathematics. Scientific calculators have the key “ex.” If you enter one for x, the calculator will display the value e.

The curve is:

f(x) = 0.25e–0.25x where x is at least zero and m = 0.25.

For example, f(5) = 0.25e(-0.25)(5) = 0.072. The value 0.072 is the height of the curve when x = 5. In (Figure) below, you will learn how to find probabilities using the decay parameter.

The graph is as follows:

Exponential graph with increments of 2 from 0-20 on the x-axis of μ = 4 and increments of 0.05 from 0.05-0.25 on the y-axis of m = 0.25. The curved line begins at the top at point (0, 0.25) and curves down to point (20, 0). The x-axis is equal to a continuous random variable.

Notice the graph is a declining curve. When x = 0,

f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m. The maximum value on the y-axis is m.

Try It

The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.

a. Using the information in (Figure), find the probability that a clerk spends four to five minutes with a randomly selected customer.

a. Find P(4 < x < 5).

The cumulative distribution function (CDF) gives the area to the left.

P(x < x) = 1 – e–mx

P(x < 5) = 1 – e(−0.25)(5) = 0.7135 and P(x < 4) = 1 – e(–0.25)(4) = 0.6321

Exponential graph with the curved line beginning at point (0, 0.25) and curves down towards point (∞, 0). Two vertical upward lines extend from points 4 and 5 to the curved line. The probability is in the area between points 4 and 5.

NOTE

You can do these calculations easily on a calculator.

The probability that a postal clerk spends four to five minutes with a randomly selected customer is P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814.

On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).

b. Half of all customers are finished within how long? (Find the 50th percentile)

b. Find the 50th percentile.

Exponential graph with the curved line beginning at point (0, 0.25) and curves down towards point (∞, 0). A vertical upward line extends from point k to the curved line. The probability area from 0-k is equal to 0.50.

P(x < k) = 0.50, k = 2.8 minutes (calculator or computer)

Half of all customers are finished within 2.8 minutes.

You can also do the calculation as follows:

P(x < k) = 0.50 and P(x < k) = 1 –e–0.25k

Therefore, 0.50 = 1 − e−0.25k and e−0.25k = 1 − 0.50 = 0.5

Take natural logs: ln(e–0.25k) = ln(0.50). So, –0.25k = ln(0.50)

Solve for k:k=\frac{ln\left(0.50\right)}{-0.25}=2.8 minutes. The calculator simplifies the calculation for percentile k. See the following two notes.

Note

A formula for the percentile k is k=\frac{ln\left(1-AreaToTheLeft\right)}{-m} where ln is the natural log.

On the home screen, enter ln(1 – 0.50)/–0.25. Press the (-) for the negative.

c. Which is larger, the mean or the median?

c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.

Answer 2

Answer:

The 60% chances that a light bulb will last is 9.19 months.

Step-by-step explanation:

Given the lifetime of a light bulb is exponentially distributed.

The mean life of light bulb is 18 months.

Thus, the decay parameter of the light bulb is $m = 1/18$

Let T be the life time of a light bulb.

Therefore the life time can be written as  $T \sim exp(1/18)$.

Thus, the probability of cumulative distribution function is given by

$P\left(T < t\right)=1-{e}^{-t/18}}$

Therefore,

$P(T < 1) = 1 - e^{-{1}/18}$

               $=1 - e^{-0.056}=1-0.945= 0.055$

Given that, there are 60% or 0.60 chances that a light bulb will last at most t months. Thus  

$P\left(T > t\right)=1-(1-{e}^{-t/18}})={e}^{-{t/18}$

Solving for the value of t,

$e^{-{t/18}} = 0.60$

$\implies -\dfrac{t}{18} = ln(0.60)$

$\implies t = -18ln(0.60) = -18\times (-0.5108)=9.19\mbox{months}$

Therefore, the 60% chances that a light bulb will last is 9.19 months.

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