The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution of this parti

Question

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution of this particular variable is very right skewed. (a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

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Serenity 1 week 2021-11-16T03:50:13+00:00 1 Answer 0 views 0

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    2021-11-16T03:52:03+00:00

    Answer:

    33.36% probability that X is less than 2.

    Step-by-step explanation:

    The distribution is not normal, however, using the central limit theorem, it is going to be approximately normal. So

    Central Limit Theorem:

    The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

    Normal Probability distribution:

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 2.2, \sigma = 1.4, n = 9, s = \frac{1.4}{\sqrt{9}} = 0.467

    (a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

    This is the pvalue of Z when X = 2. So

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{2 - 2.2}{0.467}

    Z = -0.43

    Z = -0.43 has a pvalue of 0.3336.

    33.36% probability that X is less than 2.

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45:7+7-4:2-5:5*4+35:2 =? ( )