In Exercises 19 and 20, find the mean, variance, and standard ...

Question

In Exercises 19 and 20, find the mean, variance, and standard deviation of the binomial distribution for the given random variable. Interpret the results and determine any unusual values.

About 13% of U.S. drivers are uninsured. You randomly select eight U.S. drivers and ask them whether they are uninsured. The random variable represents the number who are uninsured. (Source: Insurance Research Council)

Accepted Answer

Welcome back, everyone. A transportation survey reports that 13% of daily commuters regularly ride bicycles to work. You randomly choose 8 commuters and ask whether they ride a bicycle to work. Let the random vaer represent the number of commuters who do this. First, find the mean variance and standard deviation of this binomial distribution. Second, interpret these results in context. And third, identify any unusual values of the random variable. And explain your reasoning. Now, in this problem, notice that we are given a binomial distribution, OK, because it's either they ride the bike or not to work. And in this binomial distribution, so given a binomial distribution. With the number of trials n equal to it. The probability of success, the probability they ride a bike to work is 0.13, OK? And the probability they don't ride a bike to work, that is the probability of failure is 1 minus 0.13 or 0.87. Then we can use what we know about probabilities to find mean variance and standard deviation. Now starting with the mean, recall that the mean will be equal to the number of trials multiplied by the probability. In this case, that's 8 multiplied by 0.13, which equals 1.04. In other words, 1.04 commuters on average ride their bikes to work. That's what that means, OK? Or are expected to ride their bicycles to work. Next, when it comes to the standard devi sorry, the variants. Recall that the variance, OK, which is sigma squared, let me write that properly here. Sigma squared is equal to N multiplied by P multiplied by Q, which is 8 multiplied by 0.13 multiplied by 0.87. That's approximately 0.90. I know this means then that the standard deviation. Sigma It's going to be the square root of the variance, square root of 0.90, which is approximately 0.95. So this tells us then. That our mean is 1.04, variance 0.9, and standard deviation 0.95. Now, let's interpret these results in context. What do they mean? Well, like we said earlier, on average, when it comes to the mean, this tells us that about 0, 1.04 commuters. Out of it. Are expected to ride their bicycles to work. OK. And no, remember that variance and standard deviation tell us about the spread. So with a spread of 0.95 compared to an average of 1.04, this tells us that the spread or variability of this distribution is moderate, OK. That means it's moderate. Distribution is moderate, OK? With a standard deviation of approximately 0.95 commuters. Now let's move on to the third part of this problem, OK? Remember that we're supposed to figure out if there are any unusual values of the random variable along with an explanation for our reasoning. Now how do we know if something is unusual? Recall that value is considered unusual if it falls more than 2 standard deviations above or below the mean. So value is unusual. Generally, let's say, let's call this value X. So value X is unusual if X is less than. OK, X is less than mu minus 2 sigma or X is greater than mu + 2 sigma. OK? So it's above or below 2 standard deviations from the mean. So what would both of those values be to help us figure out what above or below them would be unusual? Well, starting with the lower bound. OK. Then, the lower bound is going to be mu minus 2 sigma. So that's 1.04 minus, so equals 1.04 minus 2 multiplied by 0.95. Which is approximately 0.86. And now, let's round that off to approximately zero since you can't have negative writers, OK? And on the other hand, in the upper bound. It's going to be mu + 22 sigma. So that's 1.04 + 2 multiplied by 0.95, which is approximately 2.94. Again, we know we can't have 2.94 of a rider, so let's round that up to 3, OK? Let's round that up to 3. So what does this tell us? Well, no, this means that any value, OK. Of X greater than or equal to 3 would be considered unusual, OK? So that means that values like 34567 or 8 are considered unusual in this context. Therefore, To summarize, the mean of our data set or the mean of our distribution is 1.04, the variance is 0.9, and the standard deviation is 0.95. On average, about 1.04 commuters out of 8 are expected to ride their bicycles to work, and we can say that the spread of dis distribution is moderate since the standard deviation is 0.95. Finally, values greater than 2 are considered unusual, OK, because they fall more than 2 standard deviations above the mean. So that's anything that's greater than or equal to 3. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Binomial Distribution

Mean, Variance, and Standard Deviation

Unusual Values

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