This is a problem of hypothesis testing where we are testing if the accident rate in Misty Glen is higher than the general rate of 9%. We can use the normal approximation to the binomial distribution to solve this problem.

Step 1: Calculate the mean and standard deviation of the binomial distribution.
The mean (μ) of a binomial distribution is n*p, where n is the number of trials and p is the probability of success. In this case, n=200 and p=0.09, so μ=200*0.09=18.
The standard deviation (σ) of a binomial distribution is sqrt(n*p*(1-p)). In this case, σ=sqrt(200*0.09*(1-0.09))=4.03.

Step 2: Calculate the z-score for the observed number of accidents.
The z-score is (X-μ)/σ, where X is the observed number of successes. In this case, X=30, so the z-score is (30-18)/4.03=2.98.

Step 3: Look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with this z-score. The probability associated with a z-score of 2.98 is 0.9986. However, this is the probability of observing 30 or fewer accidents. We want the probability of observing 30 or more accidents, so we need to subtract this probability from 1.

1 - 0.9986 = 0.0014

So, the probability of observing 30 or more car accidents in a random sample of 200 drivers, if the true accident rate is 9%, is approximately 0.0014 or 0.14%.

This is a very small probability, which suggests that it is unlikely to observe this many accidents if the true accident rate is 9%. Therefore, there is evidence to suggest that the accident rate in Misty Glen is higher than 9%.

So, the correct answer is 0.002; yes. Suggestion is that the rate is actually higher for Misty Glen. (Note: The exact probability may vary slightly depending on rounding and the method used to calculate it.)

Question

This is a problem of hypothesis testing where we are testing if the accident rate in Misty Glen is higher than the general rate of 9%. We can use the normal approximation to the binomial distribution to solve this problem.

Step 1: Calculate the mean and standard deviation of the binomial distribution.
The mean (μ) of a binomial distribution is n*p, where n is the number of trials and p is the probability of success. In this case, n=200 and p=0.09, so μ=200*0.09=18.
The standard deviation (σ) of a binomial distribution is sqrt(n*p*(1-p)). In this case, σ=sqrt(200*0.09*(1-0.09))=4.03.

Step 2: Calculate the z-score for the observed number of accidents.
The z-score is (X-μ)/σ, where X is the observed number of successes. In this case, X=30, so the z-score is (30-18)/4.03=2.98.

Step 3: Look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with this z-score. The probability associated with a z-score of 2.98 is 0.9986. However, this is the probability of observing 30 or fewer accidents. We want the probability of observing 30 or more accidents, so we need to subtract this probability from 1.

1 - 0.9986 = 0.0014

So, the probability of observing 30 or more car accidents in a random sample of 200 drivers, if the true accident rate is 9%, is approximately 0.0014 or 0.14%.

This is a very small probability, which suggests that it is unlikely to observe this many accidents if the true accident rate is 9%. Therefore, there is evidence to suggest that the accident rate in Misty Glen is higher than 9%.

So, the correct answer is 0.002; yes. Suggestion is that the rate is actually higher for Misty Glen. (Note: The exact probability may vary slightly depending on rounding and the method used to calculate it.)

Knowee AI · Accepted Answer

This is a problem of hypothesis testing where we are testing if the accident rate in Misty Glen is higher than the general rate of 9%. We can use the normal approximation to the binomial distribution to solve this problem.

Step 1: Calculate the mean and standard deviation of the binomial distribution.
The mean (μ) of a binomial distribution is n*p, where n is the number of trials and p is the probability of success. In this case, n=200 and p=0.09, so μ=200*0.09=18.
The standard deviation (σ) of a binomial distribution is sqrt(n*p*(1-p)). In this case, σ=sqrt(200*0.09*(1-0.09))=4.03.

Step 2: Calculate the z-score for the observed number of accidents.
The z-score is (X-μ)/σ, where X is the observed number of successes. In this case, X=30, so the z-score is (30-18)/4.03=2.98.

Step 3: Look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with this z-score. The probability associated with a z-score of 2.98 is 0.9986. However, this is the probability of observing 30 or fewer accidents. We want the probability of observing 30 or more accidents, so we need to subtract this probability from 1.

1 - 0.9986 = 0.0014

So, the probability of observing 30 or more car accidents in a random sample of 200 drivers, if the true accident rate is 9%, is approximately 0.0014 or 0.14%.

This is a very small probability, which suggests that it is unlikely to observe this many accidents if the true accident rate is 9%. Therefore, there is evidence to suggest that the accident rate in Misty Glen is higher than 9%.

So, the correct answer is 0.002; yes. Suggestion is that the rate is actually higher for Misty Glen. (Note: The exact probability may vary slightly depending on rounding and the method used to calculate it.)

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