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dolphi86 [110]
3 years ago
8

Over the past four years, a stock produced returns of 13 percent, −9 percent, 8 percent, and 14 percent, respectively. Based on

these four years, what range of returns would you expect to see 99 percent of the time?
Business
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

The range of returns that would be expected to be seen 99 percent of the time is (-25.5%, 38.5%).

Explanation:

Let the random variable <em>X</em> represent the percentage of returns produced by a stock.

The data for the return produced over the past four year is:

S = {13%, -9%, 8%, 14%}

Compute the average return as follows:

\text{Average Return}=\bar X

                         =\frac{1}{4}\times [0.13-0.09+0.08+0.14]\\\\=0.065

Compute the standard deviation of returns as follows:

\text{Standard deviation of Returns}(s)=\sqrt{\frac{1}{n-1}\cdot\ \sum(X-\bar X)^{2}}\\\\

=\sqrt{\frac{1}{4-1}\cdot\ [(0.13-0.065)^{2}+(-0.09-0.065)^{2}\\+(0.08-0.065)^{2}+(0.14-0.065)^{2}]}\\\\=0.1066

The 99% probability range is given by:

\text{99 percent range}=\bar X\pm 3\cdot s

Compute the range of returns that would be expected to be seen 99 percent of the time as follows:

\text{99 percent range}=\bar X\pm 3\cdot s

                          =0.065\pm 3\times 0.1066\\\\=0.065\pm 0.3198\\\\=(-0.2548,0.3848)\\\\\approx (-0.255,0.385)

Thus, the range of returns that would be expected to be seen 99 percent of the time is (-25.5%, 38.5%).

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Suppose the mean income of firms in the industry for a year is 75 million dollars with a standard deviation of 17 million dollar
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Answer and Explanation:

Given:

μ = 75 million

SD = 17 million

Probability (x) raw data = 110 million

Computation:

= Probability (x) < 110 million

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[(x-μ) / SD] = Z

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= Probability [z] < [2.05882353]

Using z calculator:

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