document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1194265>Question 1194265</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college<BR>\n" );
document.write( "students. She believes 45 percent of college students watch soap operas during the afternoon. To further investigate, she selects a sample of 10.<BR>\n" );
document.write( "a. Develop a probability distribution for the number of students in the sample who watch<BR>\n" );
document.write( "soap operas.<BR>\n" );
document.write( "b. Find the mean and the standard deviation of this distribution.<BR>\n" );
document.write( "c. What is the probability of finding exactly four watch soap operas?<BR>\n" );
document.write( "d. What is the probability less than half of the students selected watch soap operas?<BR>\n" );
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document.write( "Part (a)\r<BR>\n" );
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document.write( "X = number of students who watch soap operas<BR>\n" );
document.write( "The value of X must be equal to exactly one of the following values in the set {0,1,2,3,4,5,6,7,8,9,10}\r<BR>\n" );
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document.write( "We have a binomial distribution problem here<BR>\n" );
document.write( "This is because either a person watches soap operas or they don't (two outcomes). Each person is independent of one another. We'll assume that the ten people are isolated so they don't affect one another. <BR>\n" );
document.write( "The probability of success is p = 0.45 due to the 45% of college students who watch soap operas. The sample size is n = 10.\r<BR>\n" );
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document.write( "The binomial probability formula is<BR>\n" );
document.write( "P(X) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "where the n C x refers to the nCr combination formula. You could use Pascal's Triangle to compute nCr values.\r<BR>\n" );
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document.write( "Let's find the probability of exactly 0 people watching soap operas. So we want to find the P(X) value when X = 0.\r<BR>\n" );
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document.write( "P(X) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "P(X) = (10 C x)*(0.45)^x*(1-0.45)^(10-x)<BR>\n" );
document.write( "P(0) = (10 C 0)*(0.45)^0*(1-0.45)^(10-0)<BR>\n" );
document.write( "P(0) = (1)*(0.45)^0*(1-0.45)^(10-0)<BR>\n" );
document.write( "P(0) = 0.002533<BR>\n" );
document.write( "The approximate probability of selecting exactly 0 people who watch soap operas is about 0.002533 (i.e. roughly a 0.2533% chance)\r<BR>\n" );
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document.write( "Repeat these steps for X = 1<BR>\n" );
document.write( "P(X) = (n C x)*(p)^x*(1-p)^(n-x)<BR>\n" );
document.write( "P(X) = (10 C x)*(0.45)^x*(1-0.45)^(10-x)<BR>\n" );
document.write( "P(1) = (10 C 1)*(0.45)^0*(1-0.45)^(10-0)<BR>\n" );
document.write( "P(1) = (10)*(0.45)^1*(1-0.45)^(10-1)<BR>\n" );
document.write( "P(1) = 0.020724<BR>\n" );
document.write( "The approximate probability of selecting exactly 1 person who watches soap operas is about 0.020724 (i.e. roughly a 2.0724% chance)\r<BR>\n" );
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document.write( "If you kept going (X = 2, X = 3, ... all the way up to X = 10), then you should have this approximate probability distribution<BR>\n" );
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document.write( "<tr><TD >X</TD><TD >P(X)</TD></tr><tr><TD >0</TD><TD >0.002533</TD></tr><tr><TD >1</TD><TD >0.020724</TD></tr><tr><TD >2</TD><TD >0.076303</TD></tr><tr><TD >3</TD><TD >0.166478</TD></tr><tr><TD >4</TD><TD >0.238367</TD></tr><tr><TD >5</TD><TD >0.234033</TD></tr><tr><TD >6</TD><TD >0.159568</TD></tr><tr><TD >7</TD><TD >0.074603</TD></tr><tr><TD >8</TD><TD >0.02289</TD></tr><tr><TD >9</TD><TD >0.004162</TD></tr><tr><TD >10</TD><TD >0.000341</TD></tr>\n" );
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document.write( "Use of spreadsheet software is strongly recommended.<BR>\n" );
document.write( "I used LibreOffice which is a free program, though another useful tool is Google Spreadsheets which is also free. There are likely tons of other options to pick from. <BR>\n" );
document.write( "Feel free to use a TI83 or TI84 (or other calculator model) if you prefer that route better.\r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "mu = mean of the binomial distribution<BR>\n" );
document.write( "mu = n*p<BR>\n" );
document.write( "mu = 10*0.45<BR>\n" );
document.write( "mu = 4.5<BR>\n" );
document.write( "The mean is 4.5\r<BR>\n" );
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document.write( "sigma = standard deviation of the binomial distribution<BR>\n" );
document.write( "sigma = sqrt(n*p*(1-p))<BR>\n" );
document.write( "sigma = sqrt(10*0.45*(1-0.45))<BR>\n" );
document.write( "sigma = 1.573213<BR>\n" );
document.write( "The standard deviation is roughly 1.573213\r<BR>\n" );
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document.write( "Part (c)\r<BR>\n" );
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document.write( "Refer to the probability distribution table made in part (a)\r<BR>\n" );
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document.write( "Locate the row that has X = 4 and its corresponding probability is roughly P(X) = 0.238367\r<BR>\n" );
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document.write( "There's about a 23.8367% chance of selecting exactly 4 people (in any order) who watch soap operas.\r<BR>\n" );
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document.write( "Part (d)\r<BR>\n" );
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document.write( "Once again, refer to the table made in part (a).\r<BR>\n" );
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document.write( "Add up the P(X) values that correspond to X = 0 through to X = 4. <BR>\n" );
document.write( "Half of the sample size n = 10 is n/2 = 10/2 = 5. <BR>\n" );
document.write( "We want strictly less than this, meaning that X = 4 is the highest we can go.\r<BR>\n" );
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document.write( "P(X < 5) = P(X ≤ 4)<BR>\n" );
document.write( "P(X < 5) = P(0) + P(1) + P(2) + P(3) + P(4) <BR>\n" );
document.write( "P(X < 5) = 0.002533 + 0.020724 + 0.076303 + 0.166478 + 0.238367 <BR>\n" );
document.write( "P(X < 5) = 0.504405\r<BR>\n" );
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document.write( "The probability of less than half the students watching a soap opera is about 0.504405<BR>\n" );
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