Question

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The p.d.f that characterizes the proportion Y that make a profit is given by: ky^4(1-y)^3 and 0 elsewhere.
A. What is the value of k that renders the above a valid density function?
B.Find the probability that at most 50% of the firms make a profit in the first year.
C. Find the probability that at least 80% of the firms make a profit in the first year.

Answer 1

Answer On Question#70363 – Math – Statistics and Probability

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The p.d.f that characterizes the proportion Y that make a profit is given by: $ky^4 (1 - y)^3$ and 0 elsewhere.

A. What is the value of $k$ that renders the above a valid density function?

B. Find the probability that at most $50\%$ of the firms make a profit in the first year.

C. Find the probability that at least $80\%$ of the firms make a profit in the first year.

Solution.

A.

$$\begin{array}{l} \int_{0}^{1} k y^{4} (1 - y)^{3} d y = k \int_{0}^{1} \left(y^{4} - 3 y^{5} + 3 y^{6} - y^{7}\right) d y = k \left(\frac{y^{5}}{5} - \frac{3 y^{6}}{6} + \frac{3 y^{7}}{7} - \frac{y^{8}}{8}\right) \Bigg|_{0}^{1} = \\ = \frac{k}{280} \Rightarrow k = 280. \end{array}$$

B.

$$\begin{array}{l} \int_{0}^{0.5} 280 y^{4} (1 - y)^{3} d y = 280 \int_{0}^{0.5} \left(y^{4} - 3 y^{5} + 3 y^{6} - y^{7}\right) d y = \\ = 280 \left(\frac{y^{5}}{5} - \frac{3 y^{6}}{6} + \frac{3 y^{7}}{7} - \frac{y^{8}}{8}\right) \Bigg|_{0}^{0.5} \cong 0.363. \end{array}$$

C.

$$\begin{array}{l} \int_{0.8}^{1} 280 y^{4} (1 - y)^{3} d y = \\ = 280 \int_{0.8}^{1} \left(y^{4} - 3 y^{5} + 3 y^{6} - y^{7}\right) d y = \\ = 280 \left(\frac{y^{5}}{5} - \frac{3 y^{6}}{6} + \frac{3 y^{7}}{7} - \frac{y^{8}}{8}\right) \Bigg|_{0.8}^{1} \cong 0.056. \end{array}$$

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