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 of the drug \(-\) that is, the length of time that a company has to recover research and development costs and make a profit. Suppose the distribution of the lengths of patent life for new drugs is as shown here: $$\begin{array}{l|cccccc}\text { Years, } x & 3 & 4 & 5 & 6 & 7 & 8 \\\\\hline p(x) & .03 & .05 & .07 & .10 & .14 & .20 \\\\\text { Years, } x & 9 & 10 & 11 & 12 & 13 & \\\\\hline p(x) & .18 & .12 & .07 & .03 & .01 &\end{array}$$ a. Find the expected number of years of patent life for a new drug. b. Find the standard deviation of \(x\). c. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

Short answer

Based on the given discrete probability distribution for the number of years of patent life for new drugs, the expected number of years of patent life is 7.54 years, and its standard deviation is 1.778 years. The probability that the patent life falls into the interval of the mean plus or minus two standard deviations (\(\mu \pm 2\sigma\)) is 0.96 or 96%.

Step-by-step solution

Calculate the expected value, \(\mu\)

To calculate the expected value, multiply each value of \(x\) by its corresponding probability and sum up the results: $$\mu = E(x) = \sum_{i=1}^{n} x_i p(x_i)$$ where \(x_i\) are the years and \(p(x_i)\) are their respective probabilities.

Calculate the variance, \(\sigma^2\)

To calculate the variance, first find the expected value of the squared variable, \(E(x^2)\). Then, the variance is: $$\sigma^2 = E(x^2) - \mu^2$$

Calculate the standard deviation, \(\sigma\)

The standard deviation is equal to the square root of variance: $$\sigma = \sqrt{\sigma^2}$$

Calculate the probability for the interval \(\mu \pm 2\sigma\)

To find the probability that x falls into the interval \(\mu \pm 2\sigma\), calculate the cumulative probability of the interval and then subtract the lower bound: $$P(\mu-2\sigma\leq x \leq\mu+2\sigma) = P(x \leq\mu+2\sigma) - P(x\leq\mu-2\sigma)$$ Now let's apply these steps:

Calculate the expected value, \(\mu\)

$$\mu = E(x) = 3(0.03) + 4(0.05) + 5(0.07) + 6(0.10) + 7(0.14) + 8(0.20) + 9(0.18) + 10(0.12) + 11(0.07) + 12(0.03) + 13(0.01) = 7.54$$ The expected number of years of patent life for a new drug is 7.54 years.

Calculate the variance, \(\sigma^2\)

First, let's find the expected value of \(x^2\): $$E(x^2) = 3^2(0.03) + 4^2(0.05) + 5^2(0.07) + 6^2(0.10) + 7^2(0.14) + 8^2(0.20) + 9^2(0.18) + 10^2(0.12) + 11^2(0.07) + 12^2(0.03) + 13^2(0.01) = 57.61$$ Now we calculate the variance: $$\sigma^2 = E(x^2) - \mu^2 = 57.61 - (7.54)^2 = 3.1596$$

Calculate the standard deviation, \(\sigma\)

$$\sigma = \sqrt{\sigma^2} = \sqrt{3.1596} = 1.778$$ The standard deviation of \(x\) is 1.778 years.

Calculate the probability for the interval \(\mu \pm 2\sigma\)

First, let's find the interval \(\mu \pm 2\sigma\): $$\mu-2\sigma = 7.54 - 2(1.778) = 3.984$$ $$\mu+2\sigma = 7.54 + 2(1.778) = 11.096$$ Now, calculate the cumulative probabilities: $$P(x \leq 3.984) = P(x\leq 3) = 0.03$$ $$P(x \leq 11.096) = P(x\leq 11) = 0.03 + 0.05 + 0.07 + 0.10 + 0.14 + 0.20 + 0.18 + 0.12 + 0.07 = 0.99$$ Finally, $$P(\mu-2\sigma\leq x \leq\mu+2\sigma) = P(x \leq\mu+2\sigma) - P(x\leq\mu-2\sigma) = 0.99 - 0.03 = 0.96$$ The probability that \(x\) falls into the interval \(\mu \pm 2\sigma\) is 0.96 or 96%.

Key concepts

Expected Value

The concept of expected value is fundamental in probability distributions. It represents the average outcome you would expect after many trials of a random event. Essentially, it's a weighted average, where each possible outcome is weighted by its probability of occurrence.

In the context of our problem, we calculate the expected value, denoted by \( \mu \), for the patent life of a new drug. To find this, we multiply each possible patent life (in years) by its corresponding probability. Then, sum all those products together to get the expected number of years.

As shown in the solution, the expected number of years for a patent is calculated as follows:

• For each year \( x \), multiply by its probability \( p(x) \).
• Add all the values: \( \mu = (3 \times 0.03) + (4 \times 0.05) + ... + (13 \times 0.01) = 7.54 \).

The expected value of 7.54 years gives us an idea of the average patent life a company can expect for a new drug.

Variance

Variance measures how much the values of a random variable, like patent life, spread out from the expected value. It helps us understand the variability or risk associated with different outcomes.

The variance calculation involves two main steps:

• First, compute the expected value of the square of the variable, represented as \( E(x^2) \).
• Then, follow the formula: \( \sigma^2 = E(x^2) - \mu^2 \).

In our case, the variance \( \sigma^2 \) captures how patent life values deviate from the average of 7.54 years. Higher variance indicates more spread out values, whereas a lower variance means the values are closer to the expected value.

For the example provided:

• Compute \( E(x^2) = (3^2 \times 0.03) + (4^2 \times 0.05) + ... + (13^2 \times 0.01) = 57.61 \)
• Variance \( \sigma^2 = 57.61 - (7.54)^2 = 3.1596 \)

The variance of 3.1596 suggests moderate variability around the expected patent life.

Standard Deviation

Standard deviation builds upon variance, providing a measure of dispersion that is in the same units as the data values, making it more intuitive.

Since variance is expressed in squared units, standard deviation is simply the square root of the variance. This transformation makes it easier to interpret the spread of data values, such as patent life.

From our calculation, the standard deviation \( \sigma \) is derived by taking the square root of the variance:

• \( \sigma = \sqrt{3.1596} = 1.778 \) years

This means, on average, the duration a patent life varies by about 1.778 years from the expected value of 7.54 years. Planning around this variability informs companies about the uncertainty in patent life and potential risks.

Cumulative Probability

Cumulative probability refers to the probability that a random variable is less than or equal to a certain value. In practice, it helps determine the likelihood of a variable falling within a specific range.

The problem involves finding the probability that the patent life lies within the range \( \mu \pm 2\sigma \). This interval captures most of the probable outcomes of patent life due to the properties of the normal distribution. Here's how it's calculated:

• First, determine the interval range by calculating \( \mu - 2\sigma \) and \( \mu + 2\sigma \).
• Then, assess the cumulative probability up to these points.

In the example:

• Upper limit \( \mu + 2\sigma = 11.096 \) years
• Lower limit \( \mu - 2\sigma = 3.984 \) years
• Cumulative probability: \( P(\mu-2\sigma \leq x \leq \mu+2\sigma) = 0.99 - 0.03 = 0.96 \)

Hence, there's a 96% chance that a new drug's patent life will fall between roughly 4 to 11 years, indicating high confidence within this range.