Question

For the random variables described, find and graph the probability distribution for \(x .\) Then calculate the mean, variance, and standard deviation. Of adults 18 years and older, \(47 \%\) admit to texting while driving. ' Three adults are randomly selected and \(x\), the number who admit to texting while driving is recorded.

Short answer

Based on the given probability of adults texting while driving (0.47), calculate the probability distribution, mean, variance, and standard deviation of the number of adults (out of 3) that admit to texting while driving. Answer: - Probability distribution: P(X=0) = 0.148877, P(X=1) = 0.396987, P(X=2) = 0.350043, P(X=3) = 0.103823 - Mean: 1.41 - Variance: Approximately 0.7461 - Standard deviation: Approximately 0.864

Step-by-step solution

Identify the given parameters

In this exercise, we're given: - The probability of success (admitting to texting while driving), which is \(p = 0.47\) - The number of trials (number of adults selected), which is \(n = 3\)

Find probabilities for all possible values of \(x\)

For a binomial distribution, the probability of observing exactly \(x\) successes in \(n\) trials is given by the formula: \(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\) In this case, \(x\) can take on values between 0 and 3 (inclusive), as all, some, or none of the adults can admit to texting while driving. For \(x = 0:\) \(P(X=0) = \binom{3}{0} (0.47)^0 (0.53)^{3-0} = \binom{3}{0} (1) (0.53)^3=1\cdot0.148877=0.148877\) For \(x = 1:\) \(P(X=1) = \binom{3}{1} (0.47)^1 (0.53)^{3-1}=\binom{3}{1} (0.47) (0.53)^2=3\cdot0.47\cdot0.2809=0.396987\) For \(x = 2:\) \(P(X=2) = \binom{3}{2} (0.47)^2 (0.53)^{3-2}=\binom{3}{2} (0.47)^2 (0.53)=3\cdot0.2209\cdot0.53=0.350043\) For \(x = 3:\) \(P(X=3) = \binom{3}{3} (0.47)^3 (0.53)^{3-3}=\binom{3}{3} (0.47)^3 (1)=1\cdot 0.103823 =0.103823\)

Graph the probability distribution

To graph the probability distribution, create a bar chart with x-values (0, 1, 2, 3) and the corresponding probabilities we calculated in step 2: - \(P(X=0) = 0.148877\) - \(P(X=1) = 0.396987\) - \(P(X=2) = 0.350043\) - \(P(X=3) = 0.103823\)

Calculate mean, variance, and standard deviation

For a binomial distribution, the formulas for mean, variance, and standard deviation are given by: - Mean: \(\mu = n \cdot p\) - Variance: \(\sigma^2 = n \cdot p \cdot (1 - p)\) - Standard deviation: \(\sigma = \sqrt{\sigma^2}\) Using the given values of \(n = 3\) and \(p = 0.47\), we can find: Mean: \(\mu = 3 \cdot 0.47 = 1.41\) Variance: \(\sigma^2 = 3 \cdot 0.47 \cdot (1 - 0.47) =3\cdot0.47\cdot0.53\approx 0.7461\) Standard deviation: \(\sigma = \sqrt{0.7461} \approx 0.864\) In summary, the probability distribution for \(x\) is as follows: - \(P(X=0) = 0.148877\) - \(P(X=1) = 0.396987\) - \(P(X=2) = 0.350043\) - \(P(X=3) = 0.103823\) The mean of the distribution is 1.41, the variance is approximately 0.7461, and the standard deviation is approximately 0.864.

Key concepts

Binomial Distribution

A Binomial Distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials. Each trial has two possible outcomes, commonly termed as 'success' and 'failure'.

For our example, the scenario involves texting while driving, where 'success' is defined as an individual admitting to this behavior. Using the formula for the probability of exactly x successes out of n trials, called the binomial probability formula, we can compute the likelihood of each possible outcome. The formula is given as:

\[P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\]
where:

• \(\binom{n}{x}\) denotes the binomial coefficient,
• \(p\) is the probability of success on a single trial,
• \(n\) is the number of trials, and
• \(x\) is the number of successful trials.

Improvement of understanding might come from a real-world analogy or simulation, such as flipping a coin or rolling dice, related to the chance of success on each trial to support the theoretical understanding with a visual or tangible example.

Mean of a Distribution

The mean of a distribution represents the average expected outcome if an experiment is repeated many times. For a binomial distribution, the mean is simply the total number of trials multiplied by the probability of a success in a single trial.

In practical terms, if we were to select sets of three adults many, many times, the mean, denoted by \(\mu\), tells us the average number of people we would expect to admit to texting while driving per group of three. You can think of it as if each group of three people contribute a small portion to an overall picture of the behavior in question.

The formula to calculate the mean is:
\[\mu = n \cdot p\]
Using our example with \(n = 3\) and \(p = 0.47\), we find the mean to be 1.41. This means that, on average, out of every group of three adults selected, about 1 to 2 individuals may admit to texting while driving.

Variance and Standard Deviation

Variance and standard deviation are measures of how much the values in a set of data are spread out. Variance is the average of the squared differences from the mean, and standard deviation is the square root of variance.

For our binomial distribution scenario, these measurements tell us about the variability in the number of individuals who admit to texting while driving. A high variance indicates that the actual numbers vary widely around the mean (1.41 in our case), while a low variance signifies that the values are clustered closely around the mean. In a practical sense, this could affect how predictable the behavior of texting while driving is among randomly selected groups of adults.

The formulas used are:

• Variance: \(\sigma^2 = n \cdot p \cdot (1 - p)\)
• Standard Deviation: \(\sigma = \sqrt{\sigma^2}\)

Using these formulas, we've calculated a variance of approximately 0.7461 and a standard deviation of approximately 0.864, suggesting that there's a moderate spread of values around the mean. Being the square root of variance, the standard deviation offers a more comprehensible value, indicating the average distance from the mean a data point (in this case, an adult's admission to texting while driving) is likely to be.