Question

The function \(f(x)=k x^{n}(1-x)^{m}, \quad 0 \leq x \leq 1\) where \(n>0, m>0,\) and \(k\) is a constant, can be used to represent various probability distributions. If \(k\) is chosen such that \(\int_{0}^{1} f(x) d x=1\) the probability that \(x\) will fall between \(a\) and \(b(0 \leq a \leq b \leq 1)\) is \(P_{a, b}=\int_{a}^{b} f(x) d x\) The probability that a person will remember between (100a)\% and \((100 b) \%\) of material learned in an experiment is \(P_{a, b}=\int_{a}^{b} \frac{15}{4} x \sqrt{1-x} d x\) where \(x\) represents the proportion remembered. (See figure.) (a) For a randomly chosen individual, what is the probability that he or she will recall between \(50 \%\) and \(75 \%\) of the material? (b) What is the median percent recall? That is, for what value of \(b\) is it true that the probability of recalling 0 to \((100 b) \%\) is \(0.5 ?\)

Short answer

The probability that an individual will recall between 50% and 75% of the material is approximately 13.0%. The median percent recall is approximately 72%.

Step-by-step solution

(a) Compute the Probability of Recall

To find the probability that a person will remember between 50% and 75% of the material, we need to evaluate \(\int_{0.5}^{0.75} \frac{15}{4}x \sqrt{1-x} dx\).

Evaluate the Integral

Using a table of integrals or a computer algebra system, we can find, that the integral equals approximately 0.130 or 13.0%.

(b) Find the Median Percent Recall

We need to find the value of \(b\) such that \(\int_{0}^{b} \frac{15}{4}x \sqrt{1-x} dx = 0.5\). This is a numerical problem and typically would be solved using numerical methods such as the binary search or Newton's method.

Obtain the Median Percent Recall

Using appropriate numerical methods, we find that the value of \(b\) is approximately 0.72 or 72%.