Question

The maximum patent life for a new drug is 17 years. Subtracting the length of time required by the FDA for testing and approval of the drug provides the actual patent life for the drug—that is, the length of time that the company has to recover research and development costs and to make a profit. The distribution of the lengths of actual patent lives for new drugs is given below: Years, y 3 4 5 6 7 8 9 10 11 12 13 p(y) 0.03 0.05 0.07 0.10 0.14 0.20 0.18 0.12 0.07 0.03 0.01 • Find the mean patent life for a new drug. • Find the standard deviation of Y = the length of life of a randomly selected new drug. • What is the probability that the value of Y falls in the interval µ ± 2σ?

Answer 1

Answer:

a

   $\mu=E(y) = 7.9$

b

  $\sigma = 2.17$

c

 $P(4 < X < 12) = 0.96$

Step-by-step explanation:

From the question the data given is  

     y      3         4       5       6       7       8      9      10        11      12      13

    p(y)  0.03  0.05  0.07  0.10  0.14  0.20  0.18  0.12  0.07  0.03  0.01

Generally the mean is mathematically represented as

      $\mu =E(y) = \sum_{n=3}^{13} x_i * P(x_i)$

=>   $\mu =E(y) = 3 * 0.03 + 4 * 0.05 + \cdots +13 * 0.01$

=>   $\mu=E(y) = 7.9$

Generally the standard deviation is mathematically represented as

     $\sigma = \sqrt{ [var(y)]}$

Here $[var(y)]$ is the variance which is mathematically represented as

       $[var(y)] = [\sum_{n=3}^{13} x_i^2 * P(x_i)] -[ E(y)]^2$

=>     $[var(y)] =[ (3^2 * 0.03 ) + (4^2 * 0.05)+ \cdots + (13^2 * 0.01)] -[ 7.9]^2$

=>     $[var(y)] =67.14 - 62.41$

=>     $[var(y)] =4.73$

Thus  

     $\sigma = \sqrt{ 4.73}$  

      $\sigma = 2.17$

Generally the interval given is   µ ± 2σ  i.e  [  µ - 2σ ,  µ +2σ]

substituting values  

             $[7.9 - 2(2.17) ,7.9 - 2(2.17) ]$

            $[3.56 ,12.24 ]$

Now approximating this interval to a whole number

          $[4 ,12]$

Hence the probability  that the value of Y falls in the interval µ ± 2σ is mathematically represented as

     $P(4 < X < 12) = P(x__4} ) + \cdots + P(x__{12}})$

      $P(4 < X < 12) = 0.05 + \cdots +0.03$

     $P(4 < X < 12) = 0.96$