Question

There is some evidence that, in the years , 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). (See D. Horsky and P. Swyngedouw, "Does it pay to change your company's name? A stock market perspective," Marketing Science v6., pp.320-35 .) 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 0.83%, with a standard deviation of 0.11%. Suppose that this mean and standard deviation apply 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%,89% of the relative increases in stock price lie between 0.61 % and 1.05 %.

(b) According to Chebyshev's theorem, at least ?56%,75%,84%,89% of the relative increases in stock price lie between 0.555 % and 1.105 %.

(c) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately? 68%,75%,95%,99.7% of the relative increases in stock price lie between 0.61 % and 1.05 %.

(d) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 68% of the relative increases in stock price lie between

___%

and

_____%

.

Answer 1

A)

from Chebychev theorem ; k standard deviation from value contain at least (1-1/k2)*100   values

probability =P(0.61<X<1.05)=P((0.61-0.83)/0.11)<Z<(1.05-0.83)/0.11)=P(-2<Z<2)=(1-1/2^2)*100 =75%

B)

probability =P(0.555<X<1.105)=P((0.555-0.83)/0.11)<Z<(1.105-0.83)/0.11)=P(-2.5<Z<2.5)=(1-1/2.5^2)*100=84%

c)

probability =P(0.61<X<1.05)=P((0.61-0.83)/0.11)<Z<(1.05-0.83)/0.11)=P(-2<Z<2)=0.9772-0.0228=95%

d)

68% values fall within 1 standard deviation from mean. therefore

corresponding interval =(0.83-1*0.11,0.83+1*0.11)=(0.72 % and 0.94 %)