SOLUTION: Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college students. She believes 45 percent of college students watch soap operas during the after

Question 1194265: Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college
students. She believes 45 percent of college students watch soap operas during the afternoon. To further investigate, she selects a sample of 10.
a. Develop a probability distribution for the number of students in the sample who watch
soap operas.
b. Find the mean and the standard deviation of this distribution.
c. What is the probability of finding exactly four watch soap operas?
d. What is the probability less than half of the students selected watch soap operas?

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

X = number of students who watch soap operas
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}

We have a binomial distribution problem here
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.
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.

The binomial probability formula is
P(X) = (n C x)*(p)^x*(1-p)^(n-x)
where the n C x refers to the nCr combination formula. You could use Pascal's Triangle to compute nCr values.

Let's find the probability of exactly 0 people watching soap operas. So we want to find the P(X) value when X = 0.

P(X) = (n C x)*(p)^x*(1-p)^(n-x)
P(X) = (10 C x)*(0.45)^x*(1-0.45)^(10-x)
P(0) = (10 C 0)*(0.45)^0*(1-0.45)^(10-0)
P(0) = (1)*(0.45)^0*(1-0.45)^(10-0)
P(0) = 0.002533
The approximate probability of selecting exactly 0 people who watch soap operas is about 0.002533 (i.e. roughly a 0.2533% chance)

Repeat these steps for X = 1
P(X) = (n C x)*(p)^x*(1-p)^(n-x)
P(X) = (10 C x)*(0.45)^x*(1-0.45)^(10-x)
P(1) = (10 C 1)*(0.45)^0*(1-0.45)^(10-0)
P(1) = (10)*(0.45)^1*(1-0.45)^(10-1)
P(1) = 0.020724
The approximate probability of selecting exactly 1 person who watches soap operas is about 0.020724 (i.e. roughly a 2.0724% chance)

If you kept going (X = 2, X = 3, ... all the way up to X = 10), then you should have this approximate probability distribution

X	P(X)
0	0.002533
1	0.020724
2	0.076303
3	0.166478
4	0.238367
5	0.234033
6	0.159568
7	0.074603
8	0.02289
9	0.004162
10	0.000341

Use of spreadsheet software is strongly recommended.
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.
Feel free to use a TI83 or TI84 (or other calculator model) if you prefer that route better.

=============================================================

Part (b)

mu = mean of the binomial distribution
mu = n*p
mu = 10*0.45
mu = 4.5
The mean is 4.5

sigma = standard deviation of the binomial distribution
sigma = sqrt(n*p*(1-p))
sigma = sqrt(10*0.45*(1-0.45))
sigma = 1.573213
The standard deviation is roughly 1.573213

=============================================================

Part (c)

Refer to the probability distribution table made in part (a)

Locate the row that has X = 4 and its corresponding probability is roughly P(X) = 0.238367

There's about a 23.8367% chance of selecting exactly 4 people (in any order) who watch soap operas.

=============================================================

Part (d)

Once again, refer to the table made in part (a).

Add up the P(X) values that correspond to X = 0 through to X = 4.
Half of the sample size n = 10 is n/2 = 10/2 = 5.
We want strictly less than this, meaning that X = 4 is the highest we can go.

P(X < 5) = P(X ≤ 4)
P(X < 5) = P(0) + P(1) + P(2) + P(3) + P(4)
P(X < 5) = 0.002533 + 0.020724 + 0.076303 + 0.166478 + 0.238367
P(X < 5) = 0.504405

The probability of less than half the students watching a soap opera is about 0.504405

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