**Educational Content: Understanding the Distribution of Auto Insurance Costs****Background:**CNNBC recently reported that the mean annual cost of auto insurance is 951 dollars. We assume the standard deviation is 286 dollars, and the cost is normally distributed. We take a simple random sample of 11 auto insurance policies. **Objective:**Round your answers to 4 decimal places.**Questions:**a. **Distribution of Individual Policies (X):** What is the distribution of $ X $? - **Solution:** $ X \sim N(\mu, \sigma) $ - Where $ \mu = 951 $ and $ \sigma = 286 $b. **Distribution of the Sample Mean ($\bar{x}$):** What is the distribution of $\bar{x}$? - **Solution:** $\bar{x} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$ - Where $ n = 11 $c. **Probability for a Single Policy:** What is the probability that one randomly selected auto insurance is less than $1018? - **Solution:** Enter the probability using appropriate normal distribution calculations.d. **Probability for Sample Average:** For a simple random sample of 11 auto insurance policies, find the probability that the average cost is less than $1018. - **Solution:** Enter the probability using appropriate sampling distribution calculations.e. **Assumption of Normality:** For part d), is the assumption of normality necessary? - **Options:** No/Yes (Select one)**Additional Resources:**- For help, you can **Message the instructor**.- Once you have completed the questions, click **Submit Question** to check your answers.This exercise helps in understanding the application of normal distribution in real-world scenarios like auto insurance cost estimation.--- Note: The above exercise is designed for educational purposes to illustrate the principles of statistical distributions and sampling.

Let X be the cost of auto insurance.Given,

Answered: CNNBC recently reported that the mean…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. Click the icon to view the data table. The mean is u= girl(s). (Round to one decimal place as needed.) The standard deviation is o = e# girl(s). (Round to one decimal place as needed.) Table of numbers of girls and probabilities Number of Girls x 0 1 2 3 4 5 6 7 8 átv P(x) 0.004 0.034 0.117 0.218 0.278 0.217 0.102 0.027 0.003 D - X NiZ Next
Refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. Click the icon to view the data table. The mean is u = girl(s). (Round to one decimal place as needed.) The standard deviation is o = girl(s). (Round to one decimal place as needed.)
Private nonprofit four-year colleges charge, on average, $26,375 per year in tuition and fees. The standard deviation is $6,685. Assume the distribution is normal. Let X be the cost for a randomly selected college. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X - N( b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than 24,855 per year. c. Find the 80th percentile for this distribution. $ (Round to the nearest dollar.)
Find the number of women.
Cody took his first physics exam and scored an 80. The population mean for this exam is 70, and the standard deviation is 5. What is the probability of selecting a person with a score greater than Cody’s?
The average number of kilos of meat a person consumes in a year is 90 kilos. Assume that the standard deviation is 10 kilos and the distribution is approximately normal. If a sample of 45 individuals is selected, find the probability that the mean of the sample will be 87 kilos or greater per year. Express your answer in percent form with 2 decimal places (with percent symbol - e.g. 10.00%) type your answer...
K The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 days or longer. b. If the length of pregnancy is in the lowest 3%, then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. a. The probability that a pregnancy will last 308 days or longer is (Round to four decimal places as needed.) b. Babies who are born on or before (Round to the nearest integer as needed.) days are considered premature. .
The average number of kilos of meat a person consumes in a year is 100 kilos. Assume that the standard deviation is 11 kilos and the distribution is approximately normal. If a sample of 40 individuals is selected, find the probability that the mean of the sample will be 105 kilos or greater per year. Express your answer in percent form with 2 decimal places (with the percent symbol - e.g. 10.00%). type your answer... Next
Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean height greater than 63.0 inches. O A. 0.9918 O B. 0.0082 O C. 0.2881 O D. 0.8989 ick to select your answer. DELL
CNNBC recently reported that the mean annual cost of auto insurance is 968 dollars. Assume the standard deviation is 259 dollars, and the cost is normally distributed. You take a simple random sample of 36 auto insurance policies. Round your answers to 4 decimal places. a. What is the distribution of X? X - N( 968 259 b. What is the distribution of ? T - N( 968 c. What is the probability that one randomly selected auto insurance is less than $1015? 43.1667 0.5720 d. a simple random sample of 36 auto insurance policies, find the probability that the average cost is less than $1015. 0,8619 e. For part d), is the assumption of normal necessary? O Yes No

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