295 is the distribution of ? N 1015 N 98.3333 is the probability that one randomly selected auto insurance is less than $1079? 59 ple random sample of 9 auto insurance policies, find the probability that the average cost i $1079. art d), is the assumption of normal necessary? O No Yes

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### Understanding Mean Annual Cost of Auto Insurance

CNNBC recently reported that the mean annual cost of auto insurance is $1015. With a standard deviation of $295 and assuming a normal distribution, let's explore various related statistical queries based on a random sample of 9 auto insurance policies. Please round your answers to four decimal places.

#### a. Distribution of \( X \)

The variable \( X \) represents the annual cost of auto insurance for an individual policy. Given the mean and standard deviation, the distribution of \( X \) is:

\[ X \sim N(1015, 295) \]

#### b. Distribution of \( \bar{X} \)

The variable \( \bar{X} \) represents the mean annual cost of auto insurance for a sample of 9 policies. The distribution of \( \bar{X} \) is:

\[ \bar{X} \sim N \left(1015, \frac{295}{\sqrt{9}} \right) \]

Simplifying further:

\[ \bar{X} \sim N(1015, 98.3333) \]

#### c. Probability of One Randomly Selected Auto Insurance Cost Being Less than $1079

To calculate the probability that one randomly selected auto insurance policy costs less than $1079:

\[ P(X < 1079) \]

Using normal distribution tables or a calculator, this probability is:

\[ P(X < 1079) = 0.5859 \]

#### d. Probability that the Average Cost for a Sample of 9 Policies is Less than $1079

To find the probability that the average cost (\( \bar{X} \)) of a simple random sample of 9 auto insurance policies is less than $1079:

\[ P(\bar{X} < 1079) \]

\[ P\left(Z < \frac{1079 - 1015}{\frac{295}{\sqrt{9}}}\right) \]

This exact probability value must be calculated using normal distribution tables or an appropriate statistical tool.

#### e. Assumption of Normality

For part d, it is necessary to assume normality of the distribution to apply the Central Limit Theorem appropriate when calculating the distribution of sample means:

Is the assumption of normality necessary?

\[ \text{Yes} \]

This assumption helps ensure that the sample mean distribution approximates a normal distribution, making the probability calculations valid.
Transcribed Image Text:### Understanding Mean Annual Cost of Auto Insurance CNNBC recently reported that the mean annual cost of auto insurance is $1015. With a standard deviation of $295 and assuming a normal distribution, let's explore various related statistical queries based on a random sample of 9 auto insurance policies. Please round your answers to four decimal places. #### a. Distribution of \( X \) The variable \( X \) represents the annual cost of auto insurance for an individual policy. Given the mean and standard deviation, the distribution of \( X \) is: \[ X \sim N(1015, 295) \] #### b. Distribution of \( \bar{X} \) The variable \( \bar{X} \) represents the mean annual cost of auto insurance for a sample of 9 policies. The distribution of \( \bar{X} \) is: \[ \bar{X} \sim N \left(1015, \frac{295}{\sqrt{9}} \right) \] Simplifying further: \[ \bar{X} \sim N(1015, 98.3333) \] #### c. Probability of One Randomly Selected Auto Insurance Cost Being Less than $1079 To calculate the probability that one randomly selected auto insurance policy costs less than $1079: \[ P(X < 1079) \] Using normal distribution tables or a calculator, this probability is: \[ P(X < 1079) = 0.5859 \] #### d. Probability that the Average Cost for a Sample of 9 Policies is Less than $1079 To find the probability that the average cost (\( \bar{X} \)) of a simple random sample of 9 auto insurance policies is less than $1079: \[ P(\bar{X} < 1079) \] \[ P\left(Z < \frac{1079 - 1015}{\frac{295}{\sqrt{9}}}\right) \] This exact probability value must be calculated using normal distribution tables or an appropriate statistical tool. #### e. Assumption of Normality For part d, it is necessary to assume normality of the distribution to apply the Central Limit Theorem appropriate when calculating the distribution of sample means: Is the assumption of normality necessary? \[ \text{Yes} \] This assumption helps ensure that the sample mean distribution approximates a normal distribution, making the probability calculations valid.
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