Please help

The table summarizes results from pedestrian deaths that were caused by automobile accidents.
Pedestrian Deaths
Driver
Intoxicated?
Yes
No
Pedestrian Intoxicated?
Yes
No
51
83
263
592
If two different pedestrian deaths are randomly selected, find the probability that they both involved
pedestrians that were intoxicated.
Report the answer rounded to four decimal place accuracy.

Please Help The Table Summarizes Results From Pedestrian Deaths That Were Caused By Automobile Accidents.Pedestrian

Answers

Answer 1

Answer: 0.051

Explanation:

In order to find the probability that two different pedestrian deaths were both involved in pedestrians that were intoxicated, we need to find the probability of this specific event occurring out of all possible events. To do this, we can use the formula:

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)

First, we need to find the number of pedestrian deaths that involved intoxicated pedestrians. We can see from the table that 51 pedestrian deaths were caused by an intoxicated driver and a pedestrian who was also intoxicated.

Next, we need to find the total number of possible outcomes. We can see from the table that there are a total of 51 + 83 + 263 + 592 = 999 pedestrian deaths.

Therefore, the probability that two different pedestrian deaths were both involved in pedestrians that were intoxicated is:

Probability = 51 / 999 = 0.051

Rounded to four decimal places, the probability is 0.0510


Related Questions

Heights of 100 students were measured. The 95% confidence for Their mean heights was (169.61-170.39), and the
standard deviation = 2. Then the mean of this dataset equals?
(What is the principle to solve the question؟)

Answers

The principle is that to find the mean of the dataset, one would have to measure the height of all students and then calculate the mean.

How to calculate the mean?

The mean of the dataset is not provided, but rather a confidence interval for the mean is given. The confidence interval for the mean is calculated using the sample mean, sample standard deviation, and sample size, and it provides a range of values within which we can be certain the true population mean falls with a certain level of confidence (in this case, 95%).

Therefore, the principle is that to find the mean of the dataset, one would have to measure the height of all students and then calculate the mean. With the sample mean, sample standard deviation, and sample size, one can only calculate the confidence interval of the mean.

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Can you guys help me with 4 b)?

Answers

It’s b lol reeeeeeeeeeeeee

The probability that at least 20 of the novels have fewer than 400 pages is given as follows:

0.0823 = 8.23%.

How to obtain probabilities using the normal distribution?

The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].

The parameters for the binomial distribution are given as follows:

p = 0.3, n = 50.

Hence the mean and the standard deviation for the approximation are given as follows:

[tex]\mu = 0.3 \times 50 = 15[/tex][tex]\sigma = \sqrt{0.3 \times 0.7 \times 50} = 3.24[/tex]

Using continuity correction, the probability that at least 20 of the novels have fewer than 400 pages is P(x > 19.5), which is one subtracted by the p-value of Z when X = 19.5, hence:

Z = (19.5 - 15)/3.24

Z = 1.39

Z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823 = 8.23%.

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In a city library, the mean number of pages in a novel is 525 with a standard deviation of 200.
Furthermore, 30% of the novels have fewer than 400 pages. Suppose that you randomly select 50 nov from the library.

What is the probability that at least 20 of the novels have fewer than 400 pages?

Answers

The probability that at least 20 of the novels have fewer than 400 pages is 0.086.

To find this probability, we can use the cumulative distribution function (CDF) of a binomial distribution. The binomial distribution models the number of successful outcomes in a fixed number of independent trials, where each trial has a probability of success.

In this case, the number of trials is 50, the number of successful outcomes is the number of novels with fewer than 400 pages, and the probability of success is 0.3 (30% of the novels have fewer than 400 pages).

The CDF of the binomial distribution gives the probability of having at most a certain number of successful outcomes. To find the probability of having at least 20 successful outcomes, we can subtract the probability of having less than 20 successful outcomes from 1:

[tex]P(X > = 20) = 1 - P(X < 20)[/tex]

Where X is the number of successful outcomes.

Using a binomial calculator, we can find that  [tex]P(X > = 20) = 0.086[/tex]

So the probability that at least 20 of the novels have fewer than 400 pages is 0.086.

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Can someone help me with this statistics with explanation please

Answers

Answer:

b

Explanation:

The outcome 0, 1, 1 has a sample mean of x-bar = (0+1+1)/3 = 2/3 = 0.667

For independent outcomes, the probability of an outcome is the product of the probabilities of each individual event, which in this case would be the probability of getting a 0, the probability of getting a 1, and the probability of getting another 1.

If we assume that the probability of getting a 0, a 1, or a 1 is the same (let's say p) the probability of getting the outcome 0, 1, 1 would be ppp = p^3

So the answer is A. x-bar = 0.667, p = p^3

Note that the question doesn't provide the probability of getting a 0, 1, or 1, so I assumed the probability is the same.

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