### Understanding Sampling and Probability in Skewed DistributionsThe shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is skewed right. However, records indicate that the mean time is 21.3 minutes, and the standard deviation is 3.1 minutes. Consider the following questions and solutions:**Click here to view the standard normal distribution table (page 1).****Click here to view the standard normal distribution table (page 2).**---#### (a) Determining Sample Size for Normal ModelTo compute probabilities regarding the sample mean using the normal model, what size sample would be required?- Choose the required sample size below: - A. The normal model cannot be used if the shape of the distribution is skewed right. - B. The sample size needs to be less than 30. - C. Any sample size could be used. - **D. The sample size needs to be greater than 30.** *(Correct Answer: D)*The correct choice is **D**, indicating that the Central Limit Theorem suggests a sample size greater than 30 is sufficient for the normal model even if the original distribution is skewed.#### (b) Calculating ProbabilityWhat is the probability that a random sample of $ n = 35 $ oil changes results in a sample mean time less than 20 minutes?- The probability is approximately \[ \square \].*(Round to four decimal places as needed.)*---This exercise highlights important considerations in using statistical models and distributions to predict outcomes based on sample data. Understanding sample size requirements and the influence of distribution shape is key for accurate statistical analysis.

Given that. X~N( μ , ?) μ=21.3 , ?=3.1 Z-score =( x - μ )/?

Answered: The shape of the distribution of the…

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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A normal distribution has a mean of 148 and a standard deviation of 5. Find the z-score for a data value of 136.
A car company says that the mean gas mileage for its luxury sedan is at least 25 miles per gallon (mpg). You believe the claim is incorrect and find that a random sample of 6 cars has a mean gas mileage of 24 mpg and a standard deviation of 2 mpg. At a = 0.05, test the company's claim. Assume the population is normally distributed. Click here to view the t-distribution table. Click here to view page 1 of the normal table. Click here to view page 2 of the normal table. Which sampling distribution should be used and why? A. Use a normal sampling distribution because the population is normal, and o is known. B. Use a normal sampling distribution because n> 30. C. Use a t-sampling distribution because the population is normal, and o is known. D. Use a normal sampling distribution because the population is normal, and o is unknown. E. Use a t-sampling distribution because the population is normal, and o is unknown. F. Use a t-sampling distribution because n 25 O C. Ho: µ# 25 Ha: u= 25 D.…
The mean is 146 and the standard deviation is 35. A score of 41 is how many z-scores below the mean?
The heights of 1000 students are approximately normally distributed with a mean of 175.5 centimeters and a standard deviation of 6.2 centimeters. Suppose 300 random samples of size 25 are drawn from this population and the means recorded to the nearest tenth of a centimeter. Complete parts (a) through (c) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. (a) Determine the mean and standard deviation of the sampling distribution of X. The mean is = 175.5 Hx (Type an integer or a decimal. Do not round.) The standard deviation is ox = 1.24 (Type an integer or a decimal. Do not round.) (b) Determine the expected number of sample means that fall between 173.2 and 176.7 centimeters inclusive. sample means (Round to the nearest whole number as needed.)
Look at this histogram of heights from the GSS, which happens to be a case where a variable is relatively close to Normally-distributed (Normal curve laid atop for reference). Although it is tough to “eyeball” standard deviations, I am going to give you a list of values that could possibly describe one standard deviation of this distribution. Only one is even in the “ballpark”. Please select it based on your knowledge of Normal distributions. a. One inch b. Four inches c. 20 inches d. 50 inches
A car company says that the mean gas mileage for its luxury sedan is at least 25 miles per gallon (mpg). You believe the claim is incorrect and find that a random sample of 6 cars has a mean gas mileage of 24 mpg and a standard deviation of 2 mpg. At a = 0.05, test the company's claim. Assume the population is normally distributed. Click here to view the t-distribution table. Click here to view page 1 of the normal table. Click here to view page 2 of the normal table. Which sampling distribution should be used and why? A. Use a normal sampling distribution because the population is normal, and o is known. B. Use a normal sampling distribution because n> 30. C. Use a t-sampling distribution because the population is normal, and o is known. D. Use a normal sampling distribution because the population is normal, and o is unknown. E. Use a t-sampling distribution because the population is normal, and o is unknown. O F. Use a t-sampling distribution because n 25 Ho: H#25 Ha: µ= 25 D. H : μ…
The working life of a certain electrical equipment is normally distributed with a mean of 180 days and a standard deviation of 4 days. For each of the following questions, construct a normal distribution curve and provide the answer. About what percent of the products last between 176 and 184 days? About what percent of the products last between 180 and 184 days? For each of the following questions, use the standard normal table and provide the answer. About what percent of the products last 172 or less days? About what percent of the products last 184 or more days?
In a normal distribution, x = 5 and 2=-1.25. This tells you that x = standard deviations to the (right or left) of the mean.
This is for statistics
Justin records the speeds of vehicles moving past their house. The speed limit is 50 km/h. The mean and standard deviation of 128 readings was 2-38.1 km/h and s=5.2 km/h. Assuming the data is normally distributed, what is the percent chance (approximately) that a person is travelling faster than 43.3 km/h? Use this Standard Normal Distribution curve to help answer the question: Answer: Answer 0.1% 0.5% 30 1.7% 4.4% 9.2% 15% 10 19.1% 19.1% 15% 05 00 05 Standard Deviation (0) no 4.4% 1.7% 0.5% 0.1% 25 3.5
The average wage of plumbers is normally distributed with a mean of $20 and a standard deviation of $5.50. Sketch the normal distribution using the mean and standard deviation.
If a set of observations is normally distributed, what percent of these differ from the mean by (a) more than 2.20? (b) less than 0.32g? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.

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