Question

assume a population of 43, 49, 52, and 59. assume that samples of size ( n = 2 ) are randomly selected with replacement from the population. listed below are the sixteen different samples. complete parts (a) through (c).
43,43 43,49 43,52 43,59 49,43 49,49 49,52 49,59
52,43 52,49 52,52 52,59 59,43 59,49 59,52 59,59

a. find the median of each of the sixteen samples, then summarize the sampling distribution of the medians in the format of a table representing the probability distribution of the distinct median values. use ascending order of the sample medians.

(type integers or simplified fractions. use ascending order of the sample medians.)

Explanation:

Step1: Calculate the median for each sample

For a sample of size \(n = 2\), the median is \(\frac{x_1 + x_2}{2}\) (since \(n=2\), the median of two - number set \(\{x_1,x_2\}\) is \(\frac{x_1 + x_2}{2}\)).

• For sample \((43,43)\): \(\frac{43 + 43}{2}=43\)
• For sample \((43,49)\): \(\frac{43+49}{2}=46\)
• For sample \((43,52)\): \(\frac{43 + 52}{2}=47.5\)
• For sample \((43,59)\): \(\frac{43+59}{2}=51\)
• For sample \((49,43)\): \(\frac{49 + 43}{2}=46\)
• For sample \((49,49)\): \(\frac{49+49}{2}=49\)
• For sample \((49,52)\): \(\frac{49 + 52}{2}=50.5\)
• For sample \((49,59)\): \(\frac{49+59}{2}=54\)
• For sample \((52,43)\): \(\frac{52 + 43}{2}=47.5\)
• For sample \((52,49)\): \(\frac{52+49}{2}=50.5\)
• For sample \((52,52)\): \(\frac{52 + 52}{2}=52\)
• For sample \((52,59)\): \(\frac{52+59}{2}=55.5\)
• For sample \((59,43)\): \(\frac{59 + 43}{2}=51\)
• For sample \((59,49)\): \(\frac{59+49}{2}=54\)
• For sample \((59,52)\): \(\frac{59 + 52}{2}=55.5\)
• For sample \((59,59)\): \(\frac{59+59}{2}=59\)

Step2: Count the frequencies and calculate probabilities

• Median \(43\): Frequency \(f_1 = 1\), Probability \(P(43)=\frac{1}{16}\)
• Median \(46\): Frequency \(f_2=2\), Probability \(P(46)=\frac{2}{16}=\frac{1}{8}\)
• Median \(47.5\): Frequency \(f_3 = 2\), Probability \(P(47.5)=\frac{2}{16}=\frac{1}{8}\)
• Median \(49\): Frequency \(f_4=1\), Probability \(P(49)=\frac{1}{16}\)
• Median \(50.5\): Frequency \(f_5 = 2\), Probability \(P(50.5)=\frac{2}{16}=\frac{1}{8}\)
• Median \(51\): Frequency \(f_6=2\), Probability \(P(51)=\frac{2}{16}=\frac{1}{8}\)
• Median \(52\): Frequency \(f_7 = 1\), Probability \(P(52)=\frac{1}{16}\)
• Median \(54\): Frequency \(f_8=2\), Probability \(P(54)=\frac{2}{16}=\frac{1}{8}\)
• Median \(55.5\): Frequency \(f_9=2\), Probability \(P(55.5)=\frac{2}{16}=\frac{1}{8}\)
• Median \(59\): Frequency \(f_{10}=1\), Probability \(P(59)=\frac{1}{16}\)

Answer:

Sample Median	Probability	Sample Median	Probability
\(46\)	\(\frac{1}{8}\)	\(52\)	\(\frac{1}{16}\)
\(47.5\)	\(\frac{1}{8}\)	\(54\)	\(\frac{1}{8}\)
\(49\)	\(\frac{1}{16}\)	\(55.5\)	\(\frac{1}{8}\)
\(50.5\)	\(\frac{1}{8}\)	\(59\)	\(\frac{1}{16}\)