Q. 8.145 - Elementary Statistics

Question

An issue of Scientific American revealed that the batting averages of major-league baseball players are normally distributed with mean \(.270\) and standard deviation \(.031\).

a. Simulate \(2000\) samples of five batting averages each.

b. Determine the sample mean and sample standard deviation of each of the \(2000\) samples.

c. For each of the \(2000\) samples, determine the observed value of the standardized version of \(\bar{x}\).

d. Obtain a histogram of the \(2000\) observations in part (c). 

e. Theoretically, what is the distribution of the standard version of \(\bar{x}\)?

f. Compare your results from parts (d) and (e).

g. For each of the \(2000\) samples, determine the observed value of the studentized version of \(\bar{x})\).

h. Obtain a histogram of the \(2000\) observations in part (g).

i. Theoretically, what is the distribution of the standard version of \(\bar{x})\) ?

j. Compare your results from parts (h) and (i).

k. Compare your histograms from parts (d) and (h). How and why do they differ?

Step-by-Step Solution

Answer

Part a. The 2000 random sample is 

Then the five batting average will be 

Part b. \(\bar{x}=0.2698\)

            \(\sigma=0.0134\)

Part c. 

Part d. 

Part a. Step 1. Given information

The number of sample \((n)\), mean \((\bar{x})\) and standard deviation \((\sigma)\) is given.

\(\bar{x}=0.27\)

\(\sigma=0.031\)

\(n=2000\times 5=10000\)

Part a. Step 2. Calculation

Generate \(2000\) samples of males using function “norminv” with \(0.031\) sample mean and \(0.27\) standard deviation in MATLAB

\(r=norminv(rand(2000,5), 0.27, 0.031)\)

After that we will get random sample \(2000\) males.

Program: 

Query:

• First, we have defined the number of samples.
• Then create \(2000\) random samples using function “norminv” with sample mean \(0.27\) and standard deviation \(0.031\).

Part b. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

The solution is \(\bar{x}=0.2698\)

                          \(\sigma=0.0134\)                      

Part c. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean-0.2698}{0.0134}\)

Program:

Query:

• First, we have defined the number of samples.
• Then generate the random samples.
• Then calculate the mean and standard deviation.
• Calculate the z-score which is standardized form of all samples.

Part d. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean-0.2698}{0.0134}\)

Program: 

Query: 

• First, we have defined the number of samples.
• Then generate the random samples.
• Then calculate the mean and standard deviation.
• Calculate the z-score which is standardized form of all samples.

• Then generate the histogram.