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cnnbc recently reported that the mean annual cost of auto insurance is 959 dollars. assume the standard deviation is 263 dollars, and the cost is normally distributed. you take a simple random sample of 37 auto insurance policies. round your answers to 4 decimal places.
a what is the probability that one randomly selected auto insurance is more than $984?
b. a simple random sample of 37 auto insurance policies, find the probability that the average cost is more than $984.


Sagot :

Using the normal distribution, the probabilities are given as follows:

a. 0.4602 = 46.02%.

b. 0.281 = 28.1%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

The parameters are given as follows:

[tex]\mu = 959, \sigma = 263, n = 37, s = \frac{263}{\sqrt{37}} = 43.24[/tex]

Item a:

The probability is one subtracted by the p-value of Z when X = 984, hence:

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

[tex]Z = \frac{984 - 959}{263}[/tex]

Z = 0.1

Z = 0.1 has a p-value of 0.5398.

1 - 0.5398 = 0.4602.

Item b:

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

By the Central Limit Theorem:

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

[tex]Z = \frac{984 - 959}{43.24}[/tex]

Z = 0.58

Z = 0.58 has a p-value of 0.7190.

1 - 0.719 = 0.281.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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