Question

There is some evidence that, in the years −198185 , 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 v. 6, pp. −32035,1987.) 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 0.80%, with a standard deviation of 0.10%. 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%, or 89% of the relative increases in stock price lie between 0.60 % and 1.00 %.

(b) According to Chebyshev's theorem, at least 56%, 75%, 84%, or 89% of the relative increases in stock price lie between 0.65 % and 0.95 %.

(c) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 68%, 75%, 95%, or 99.7% of the relative increases in stock price lie between 0.60 % and 1.00 %.

(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

Answer)

Mean = 0.8

S.d = 0.1

A)

According to chebyshev's theorem, at least 1 - (1/k^2) lies in between k s.d

Now, 0.6 = 0.8 - (0.1*2)

And 1 = 0.8 + (0.1*2)

That is 0.6 and 1 are with in 2 s.d

So, answer is 1 - (1/2^2)

= 75%

B)

0.65 = 0.8 - (1.5*0.1)

0.95 = 0.8 + (1.5*0.1)

That is answer is 1-(1.1.5^2) = 56%

C)

According to the emperical rule

If the data is normally distributed

Then 68% lies in between mean - s.d and mean + s.d

95% lies in between mean - 2*s.d and mean + 2*s.d

99.7% lies in between mean - 3*s.d and mean + 3*s.d

As 0.6 and 1 are with in 2s.d

So answer is 95%

D)

68% lies in between

0.8-0.1 and 0.8+0.1

0.7 and 0.9