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
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
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa28b18f-3e2d-4592-9d09-7b963ee49630%2Fdcb19c45-6f57-46c8-80ce-466802156790%2Fdpesi4_processed.jpeg&w=3840&q=75)
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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