Question

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean 28 weeks and standard deviation 14 weeks. (a) What is the probability that a transistor will last between 14 and 28 weeks? (Round your answer to three decimal places.) (b) What is the probability that a transistor will last at most 28 weeks? (Round your answer to three decimal places.) Is the median of the lifetime distribution less than 28? Why or why not? The median is 28, since P(X ≤ ) = .5. (c) What is the 99th percentile of the lifetime distribution? (Round your answer to the nearest whole number.) (d) Suppose the test will actually be terminated after t weeks. What value of t is such that only 0.5% of all transistors would still be operating at termination? (Round your answer to the nearest whole number.) t = weeks

Answer 1

Answer:

a The probability Needed is 0.424

b1 The probability Needed is 0.567

b2 The median of the life time distribution is less than 28 cause its probability is less than the probability of 28

c The Needed life time is 70 weeks

d The needed lifetime is t = 77 weeks

Explanation:

Let A represent the life time of the transistor in weeks and follows a gamma distribution with parameters ($\alpha,\beta$).

Looking at what we are give we can say that

                 E(A) = 28 weeks and  $\sigma_A$  = 14 weeks

Hence the mean of the gamma distribution is E(A) = $\alpha\beta ------(1)$

and the variance of the gamma distribution is V(A) = $\alpha\beta^2 ------(2)$

Looking at Equation 2

              $\alpha\beta^2 =V(A)$

              $(\alpha\beta)\beta = V(A)$

             $E(A)\beta = V(A)$

                     $\beta = \frac{V(A)}{E(A)}$

$Note:\alpha\beta = E(A)\\\ and \ \sigma_A^2 = V(A)$

                    $\beta = \frac{\sigma_A^2}{E(A)}$

                    $\beta = \frac{14^2}{28} = \frac{196}{28} = 7$

Looking at Equation 1

               E(A)  = $\alpha \beta$

                  28 = $\alpha (7)$

          =>    $\alpha = \frac{28}{7}$

         =>    $\alpha = 4$

Now considering the first question

        i.e to find the probability that a transistor will last between 14 and 28 weeks

            P(14 < A < 28) = P(A ≤28) - P(A ≤ 14)

The general formula for probability of  gamma distribution

                                   = $F[\frac{a}{\beta},\alpha ] - F[\frac{a}{\beta},\alpha ]$

                                   $= F[\frac{28}{7},4 ] - F[\frac{14}{7},4 ]$

                                   $= F(4,4) - F(2,4)$

                                   $= 0.567 - 0.143$    (This is gotten from the gamma table)

                                  $= 0.424$

Now considering the second question        

     i.e to find the probability that a transistor will last at most 28 weeks  

                           $P(A \le 28)$    =    $F[\frac{28}{7},4]$

                                                $=F(4,4)$

                                                $= 0.567$     (This is obtained from the gamma Table)

 we need to determine the median of the life time distribution less than 28

        from what we are given $P(X \le \mu ) = 0.5$ we can see that $\mu$<28 since $P(A \le 28)$  = 0.567 and this is greater than 0.5

Now considering the Third  question

       i.e to find the  $99^{th}$ percentile of the life time distribution

       Generally

        F(a : $\alpha$) =99%

        F(a : 4) = 0.99

Looking at the gamma table the tabulated values at $\alpha =4$ and the corresponding 0.99 percent value to a is 10

         Now , find the product of $\beta =7$ with a = 10 to obtain the $99^{th}$ percentile

                   $99^{th} \ percentile = a\beta$

                                            $= 10 * 7$

                                            $=70$

Now considering the Fourth  question

    i.e to determine the number of weeks such that 0.5% of all transistor will still be operating

    The first thing to do is to find the value in the incomplete gamma function with $\alpha = 4$ in such a way that

              F(a:4) = 1 - 0.5%

              F(a: 4) = 1 - 0.005

              F(x: 4)  = 0.995

So we will now obtain the 99.5 percentile

 Looking at the gamma tabulated values at $\alpha = 4$ and the corresponding 0.995 percent value for a (x depending on what you want to denoted it with) is 11

   So we would the obtain the product of $\beta =7$ with  a = 11  which is the $99.5^{th}$ percentile

                              $t = a\beta$

                                 $= 11 *7$

                                $= 77$

   

Answer 2

The probability will be:

(a) 0.4240

(b) 0.5670

(c) 70

(d) 77

According to the question,

→ $\mu = \alpha \beta$

 $28= \alpha \beta$ ...(equation 1)

→ $\sigma^2 = \alpha \beta^2$

 $14^2 = \alpha \beta^2$ ...(equation 2)

By putting "equation 1" in "equation 2", we get

→ $14^2 = \alpha \beta\times \beta$

   $14^2 = 28\times \beta$

→ $196 = 28\times \beta$

      $\beta = 7$

      $\alpha = 4$

(a) P(14 and 28)

→ $P(14< X< 28 ) = F(\frac{x}{\beta}, \alpha )- F (\frac{x}{\beta}, d )$

                             $= F(\frac{28}{7}, 4 )- F(\frac{14}{7} ,4)$

                             $= F(4,4)-F(2,4)$

By using gamma tables, we get

                             $= 0.5670-0.1430$

                             $= 0.4240$

(b) P(almost 28)

→ $P(x \leq 28) = F(\frac{x}{\beta}, \alpha )$

                    $= F (\frac{28}{7} ,4)$

                    $= F(4,4)$

                    $= 0.5670$

(c) 99 percentile

• $F(X \leq x, \alpha) = 0.99$
• $F(x, \alpha) = 0.99$
• $F(x:4)=0.99$

at x = 10,

99th percentile,

→ $x \beta = 10\times 7$

        $= 70$

(d)

• $F(x;4) = 1-0.005$
• $F(x;4) = 0.995$

x = 11

→ $t = x \beta$

     $= 11\times 7$

     $= 77$

Thus the above answers are correct.

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