The probability that a person will remember between and of material learned in an

Chapter 5, Problem 105

(choose chapter or problem)

In Exercises 105 and 106, 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

\(\int_{0}^{1} f(x) d x=1\)

then 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 100b% 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 b is 0.5 ?

Text Transcription:

f(x)=k x^n(1-x)^m, 0 leq x leq 1

n>0, m>0

int_0^1 f(x) d x=1

(0 leq a leq b leq 1)

P_a, b=int_a^b f(x) d x

P_a, b=int_a^b frac 15 4 x sqrt 1-x d x