Question

5There is some speculation that a simple name change can result in a short-term increase in the price of certain business firms' stocks (relative to the prices of similar stocks). Suppose that to test the profitability of name changes, we analyze the stock prices of a large sample of corporations shortly after they changed names, and we find that the mean relative increase in stock price was about 0.87%, with a standard deviation of 0.10%. Suppose that this mean and standard deviation apply to the population of all companies that changed names. Complete the following statements about the distribution of relative increases in stock price for all companies that changed names.(a)According to Chebyshev's theorem, at least ▼(Choose one 36,56,75,84,89) of the relative increases in stock price lie between 0.67% and 1.07%.(b)According to Chebyshev's theorem, at least 84% of the relative increases in stock price lie between ?????% and ????? %. (Round your answer to 2 decimal places.)

Answer 1

We are given the mean and standard deviation for the profitability of companies that changed their names.

$\begin{gathered} Mean=0.87\% \\ Standard\text{ }Deviation=0.10\% \end{gathered}$

Chebyshev's Theorem states that the estimated percentage of data falling within k standard deviations is equal to

$1-\frac{1}{k^2}$

To find the percentage of companies with relative stock price increases between 0.67% and 1.07%, we must first solve for k.

$\begin{gathered} k=\frac{1.07-0.87}{0.10} \\ \\ k=2 \end{gathered}$

Then we use k =2 in Chebyshev's formula.

$\begin{gathered} 1-\frac{1}{k^2} \\ \\ =1-\frac{1}{2^2} \\ \\ =1-\frac{1}{4} \\ \\ =\frac{3}{4} \\ \\ =0.75 \end{gathered}$

The answer is 0.75 or 75%.

For part b, we will use the following equation:

$0.84=1-\frac{1}{k^2}$

Solving for k, we get:

$\begin{gathered} 0.84=1-\frac{1}{k^2} \\ \\ -0.16=-\frac{1}{k^2} \\ \\ k^2=\frac{1}{0.16} \\ \\ k=\frac{1}{0.4} \\ \\ k=2.5 \end{gathered}$

This means that the boundaries are 2.5 standard deviations away from the mean.

$\begin{gathered} Lower\text{ }boundary=0.87\%-(2.5\times0.10\%) \\ Lower\text{b}oundary=0.87\operatorname{\%}-0.25\% \\ Lower\text{b}oundary=0.62\% \end{gathered}$$\begin{gathered} Upper\text{ }boundary=0.87\%+(2.5\times0.10\%) \\ Upper\text{ }boundary=0.87\%+0.25\% \\ Upper\text{ }boundary=1.12\% \end{gathered}$

The answers are 0.62% and 1.12%.