Question

The closing stock prices of $35$U.S. semiconductor manufacturers are given as follows.

$$\begin{gathered} 8.625;30.25;27.625;46.75;32.875;18.25;5;0.125;2.9375; \\ 6.875;28.25;24.25;21;1.5;30.25;71;43.5;49.25;2.5625;31; \\ 16.5;9.5;18.5;18;9;10.5;16.625;1.25;18;12.87;7;12.875; \\ 2.875;60.25;29.25 \end{gathered}$$

a. In words,${\rm X}=\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

b. i.$\overline{x}=\_\_\_\_\_$

ii.$s_x=\_\_\_\_\_$

iii.$n=\_\_\_\_\_$

c. Construct a histogram of the distribution of the averages. Start at $x=–0.0005$. Use bar widths of ten.

d. In words, describe the distribution of stock prices.

e. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five pieces

together until you have ten averages. List those ten averages.

f. Use the ten averages from part e to calculate the following.

i.$\overline{x}=\_\_\_\_\_$

ii.$s_x=\_\_\_\_\_$

g. Construct a histogram of the distribution of the averages. Start at $x=-0.0005$. Use bar widths of ten.

h. Does this histogram look like the graph in part c?

i. In one or two complete sentences, explain why the graphs either look the same or look different?

j. Based upon the theory of the central limit theorem,$X¯~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_)$

Short answer

a.$X=$the closing stock prices for U.S. semiconductor manufacturers

b$i.\$20.71;ii.\$17.31;iii.35$

c. Histogram of the distribution is drawn

d. Exponential distribution $X~\mathrm{Exp}\left(\frac{1}{20.71}\right)$

e. first ten random numbers are selected.

f. $i.20.71ii.11.14$

g. Histogram of the distribution is drawn.

h. No.

i. as the averaged values are different, histograms are different.

j.$N\left(20.71,\frac{17.31}{\sqrt{5}}\right)$

Step-by-step solution

Given information

According to the given information, the random variable $X$is defined as,

$X=$the closing stock prices for U.S. semiconductor manufacturers

Explanation (part b)

i. The mean of the random variable is given by, $\overline{x}=\frac{\sum _{}x}{n}=\frac{724.85}{35}=20.71$

ii. The standard deviation of the given case is calculated as, role="math" localid="1651841050717" $s_x=17.31$

iii. The total number of stock prices$n=35$

Explanation (part c)

The largest value in the given data is $71$.

The smallest value in the given data is $0.125$.

Since the decimal number carries four digits in the given data, we start with $x=-0.0005$

To find the starting point, $0.125-0.0005=0.1245$

To find the ending point, $71+0.0005=71.0005$

Now, calculate the width of each bar or class interval with bar widths of ten, we get

$\frac{71.0005-0.1245}{10}=7.0876$

Rounding off the above value we get, 7. This number is added with each data and exact Histogram of the distribution of the averages are given as:

Explanation (part d)

The distribution of stock prices is given in exponential distribution and it is given as,

$$\begin{gathered} X~\mathrm{Exp}\left(\frac{1}{\overline{x}}\right) \\ X~\mathrm{Exp}\left(\frac{1}{20.71}\right) \end{gathered}$$

Explanation (part e)

the number of averages selected randomly is $10$. The values are listed as follows,

$8.625;30.25;27.625;46.75;32.875;18.25;5;0.125;2.9375;6.875;$

Explanation (part f)

i. From part e, the mean of the averaged value is given as, $\overline{x}=20.17$

ii. Similarly, the standard deviation of the ten averaged value is given as,role="math" localid="1651841076286" $s_x=11.14$

Explanation (part f)

Using the ten averages from part e, we are calculating,

i. mean for ten averaged data, $\overline{x}=\frac{\sum _{}x}{n}=\frac{179.3}{10}=17.93$

ii. the standard deviation for ten averaged data,$s_x=15.69$

Explanation (part g)

From the ten averaged values, the largest value is $46.75$, the smallest value is $0.125$

Start at $x=-0.0005$, we get, starting point value as $0.125-0.0005=0.1245$

the ending point value as $46.75+0.0005=46.7505$.

Now using bar widths of ten, we get, $\frac{46.7505-0.1245}{10}=4.66$

Rounding off the above value to $5$and add this value with the given data sets to construct a histogram of the distribution.

Explanation (part h)

The histogram of part c is different from part g as the number of values in the data set is less.

Explanation (part i)

The graphs look different because the number of averaged data selected for the two subparts are different and also the largest values are different in the set. We can also try to select the averaged ten data sets in order to proceed with making the equal mean and standard deviation. But we can't get the same equality in histogram graphs as the data sets selected are different in both sets.

Explanation (part j)

Based on the central limit theorem, the random variable is expressed as,

$X~N\left(\mu ,\frac{\sigma }{\sqrt{n}}\right)$

Plugging all the known values in the above equation, we get,

$X~N\left(20.71,\frac{17.31}{\sqrt{5}}\right)$