Televisions In the Sullivan Statistics Survey, individuals...

Televisions In the Sullivan Statistics Survey, individuals were asked to disclose the number of televisions in their household. In the following probability distribution, the random variable $X$ represents the number of televisions in households. $$\begin{array}{cl} \text { Number of Televisions, } \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\ \hline 0 & 0 \\ \hline 1 & 0.161 \\ \hline 2 & 0.261 \\ \hline 3 & 0.176 \\ \hline 4 & 0.186 \\ \hline 5 & 0.116 \\ \hline 6 & 0.055 \\ \hline 7 & 0.025 \\ \hline 8 & 0.010 \\ \hline 9 & 0.010 \\ \hline \text { Source: Sullivan Statistics Survey } \end{array}$$ (a) Verify this is a discrete probability distribution. (b) Draw a probability histogram. (c) Determine and interpret the mean of the random variable $X .$ (d) Determine the standard deviation of the random variable $X .$ (e) What is the probability that a randomly selected household has three televisions? (f) What is the probability that a randomly selected household has three or four televisions? (g) What is the probability that a randomly selected household has no televisions? Would you consider this to be an impossible event?

Televisions In the Sullivan Statistics Survey, individuals were asked to disclose the number of televisions in their household. In the following probability distribution, the random variable $X$ represents the number of televisions in households. $$\begin{array}{cl}
\text { Number of Televisions, } \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\ \hline 0 & 0 \\ \hline 1 & 0.161 \\ \hline 2 & 0.261 \\ \hline 3 & 0.176 \\ \hline 4 & 0.186 \\ \hline 5 & 0.116 \\ \hline 6 & 0.055 \\ \hline 7 & 0.025 \\ \hline 8 & 0.010 \\ \hline 9 & 0.010 \\ \hline \text { Source: Sullivan Statistics Survey } \end{array}$$
(a) Verify this is a discrete probability distribution.
(b) Draw a probability histogram.
(c) Determine and interpret the mean of the random variable $X .$
(d) Determine the standard deviation of the random variable $X .$
(e) What is the probability that a randomly selected household has three televisions?
(f) What is the probability that a randomly selected household has three or four televisions?
(g) What is the probability that a randomly selected household has no televisions? Would you consider this to be an impossible event?

Key Concepts

Basic Probability Rules

Basic probability rules include the addition rule for mutually exclusive events, where the probability of either of two mutually exclusive events occurring is the sum of their probabilities. This concept is applied in scenarios where probabilities for combined events, like having three or four televisions, are computed by adding the individual probabilities. Additionally, understanding events with zero probability and their interpretation is an important aspect of probability theory.

Standard Deviation

The standard deviation is a measure of the spread or dispersion of a discrete probability distribution around its mean. It is calculated by taking the square root of the variance, which itself is the average of the squared differences between each outcome and the mean, weighted by the probabilities. This metric helps in understanding how much variability exists in the data.

Mean (Expected Value)

The mean, or expected value, of a discrete random variable is a measure of central tendency that is calculated by summing the products of each outcome and its probability. It represents the average outcome if the experiment were to be repeated many times, and it provides a central value around which the probabilities are distributed.

Probability Histogram

A probability histogram is a graphical representation of a probability distribution for a discrete random variable. It uses bars to display the probabilities associated with each outcome, making it easy to visualize the distribution's shape, central tendency, and spread. Each bar's height corresponds to the probability of the associated value of the variable.

Probability Distribution

A probability distribution for a discrete random variable lists each possible value the variable can assume along with its associated probability. Key criteria include that each probability must be between 0 and 1, and the sum of all the probabilities must equal 1. This ensures that the distribution is valid and represents a complete set of outcomes, as illustrated by the Sullivan Statistics Survey data.

Discrete Random Variable

A discrete random variable is one that can take on a countable number of distinct values. In the context of this problem, it represents the number of televisions in a household, which is counted as 0, 1, 2, etc. Discrete random variables are used in situations where the outcomes are separate and distinct values, rather than a continuous range.

Transcript

00:01 In this problem, we have some data revolving around the number of televisions in a household.

00:09 And the number of televisions is our random variable.

00:16 And some households reported no tvs all the way up through and including nine tvs.

00:28 And based on the survey, a probability distribution was created saying that the probability of finding no households, or households with zero tvs was zero, and the probability of finding households with one tv was 0 .161, and two tvs was 0 .261, and so forth.

00:58 So i'm just going to write all those down really quick, and we're done there.

01:22 Okay, so for part a, the question, is asking us to verify that this is indeed a discrete probability distribution.

01:32 So let's talk about the definition of discrete first.

01:36 Discrete random variables are going to be the outcomes that are integer in nature.

01:58 So when we are talking about the number of tvs, we can't talk about a household having two and a half tvs.

02:05 They either have one tv, or two tvs or three tvs.

02:09 There's no fractional responses that can come out of it.

02:12 So that's the first part of part a is understanding that, yes, these are discrete values.

02:20 The fact that it is a probability distribution, if this is going to be a probability distribution, then the sum of all the probabilities must equal one.

02:33 So we need to add up that second column, the probability column, and guarantee that that totals up to 1.

02:44 So i'm going to show you a fast way of doing that using the graphing calculator.

02:49 So i'm going to bring in my graph in calculator.

02:51 And as you can see, i've already put the random variables 0 through 9 into my calculator, and i have placed the corresponding probabilities into list 2.

03:01 So basically what we want to do here is to sum up list two.

03:06 So i'm going to quit out of the table mode, and i'm going to hit second stat, and i'm going to move over to the math tab, and i'm going to ask the calculator to sum up, which is number five, everything found in list two.

03:24 And sure enough, i can confirm that this is a probability distribution because the sum, of p of x is equivalent to one.

03:36 Part b is to draw a histogram, a probability histogram for what's going on here.

03:43 So we're going to scoot up a little bit.

03:49 And keep in mind when you're drawing histograms, your towers must touch.

03:59 And the vertical axis represents probabilities.

04:05 So we're going to start with 0 .05, 0 .10, and that should all be evenly spaced, 0 .15, 0 .20, and 0 .25.

04:26 And for 0 tvs, the probability was 0.

04:31 So therefore, i'm just going to place 0 down here, but i'm not going to have a tower corresponding to it, because the survey showed that no one in the survey had a household with no tvs.

04:47 For 1, it's 0 .161.

04:51 So 0 .161 is slightly past 0 .15.

04:57 So i'm going to go up to a little bit past 0 .15.

05:02 There's one tv.

05:04 And if you want to, you can declare for the reader what that tower tops out at.

05:11 So if you really want to, you can do that.

05:13 Then we need to do two.

05:15 The probability of two tvs was 0 .261.

05:20 So 0 .261 would end up just slightly bigger than 0 .25.

05:28 Whoops.

05:29 Slightly bigger than 0 .25.

05:35 And your towers should be uniform.

05:38 They should be the same width.

05:40 So we'll have 0 .261 right here.

05:48 Then for three tvs, we had 0 .176.

05:55 So 0 .176 is almost to the point 20.

06:08 For 4 tvs, it was 0 .186.

06:15 So that's got to be a little higher than 0 .176.

06:19 So we're going to go a little higher.

06:34 Now i'm going to write on both.

06:36 So this was 0 .176 and this was 0 .186.

06:42 And the next one is 0 .116, which is slightly over 0 .10.

06:58 And then for 6 tvs, it was 0 .055.

07:04 0 .055 is going to be slightly over the 0 .05.

07:08 And then the next one is 0 .025.

07:26 And then for 8 and 9, it's 0 .01.

07:31 So 0 .01 and 1.

07:45 So that is what our probability histogram is going to look like.

07:54 And if we had to describe this, we would describe this as a skewed right or positive.

08:02 Skewed distribution.

08:18 All right, for part c, you need to determine and interpret the mean of this random variable x.

08:27 So to determine the mean, the mean of a probability distribution will be equivalent to the sum of all x's times their corresponding p of x.

08:44 So we're going to go back to our chart and we're going to introduce a new column and that new column is going to be titled x times p of x so that means we are going to take the x value and multiply it by its corresponding p of x value so this would be zero then we'll take this x value time its corresponding x value or p of x value and get 0 .161.

09:20 And then we'll go to the next one, 2 times 0 .261, and you'll get 0 .522, and we'll keep the pattern going.

09:31 So here we'll get 0 .528 .744.

09:42 0 .58 .33 .175 .0 .08.

09:54 0 .0 .0 .0 .0 .0 .0.

09:54 0 .08.

09:58 And 09.

10:00 Now, i'm going to let the calculator generate that as well.

10:04 Just for the sake, we now have to then add up that column...