Question

There is some evidence that, in the years 1981-85, a simple name change resulted 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 in the more recent market (the past five years), 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 0.88%. Suppose that this mean applies to the population of all companies that changed names during the past five years. Complete the following statements about the distribution of relative increases in stock price for all companies that changed names during the past five years.

(a) According to Chebyshev's theorem, at least ________ (56% 75% 84% or 89%) of the relative increases in stock price lie within 2.5 standard deviations of the mean, 0.88%.

(b) Suppose that the distribution is bell-shaped. If approximately 68% of the relative increases in stock price lie between 0.69 % and 1.07 %, then the approximate value of the standard deviation for the distribution, according to the empirical rule, is ________.

Answer 1

Solution: (a) According to Chebyshev's theorem, at least 84% of the relative increases in stock price lie within 2.5 standard deviations of the mean, 0.88%.

Explanation:

The Chebyshev's theorem says that at least $1-\frac{1}{k^2}$ of the data lie within k standard deviations of the mean.

Therefore, we have:

$1-\frac{1}{2.5^2}$

$1-\frac{1}{6.25}$

$0.84=84\%$

(b) Suppose that the distribution is bell-shaped. If approximately 68% of the relative increases in stock price lie between 0.69% and 1.07 %, then the approximate value of the standard deviation for the distribution, according to the empirical rule, is 0.19%.

Explanation:

The normal distribution, 68% of data lies within 1 standard deviation from the mean.

We are given:

$\mu- 1 \times \sigma=0.69\%\Rightarrow 0.88\%-\sigma=0.69\%\Rightarrow \sigma=0.88\%-0.69\%=0.19\%$

$\mu+ 1 \times \sigma=1.07\%\Rightarrow 0.88\%+\sigma=1.07\%\Rightarrow \sigma=1.07\%-0.88\%=0.19\%$

Therefore, the standard deviation for the distribution is 0.19%