### Educational Website Content: Understanding Normal Distribution of Broiler Weights### Big Chickens: Weight Distribution AnalysisThe weights of broilers (commercially raised chickens) are approximately normally distributed with a mean of 1387 grams and a standard deviation of 192 grams. Use the TI-84 Plus calculator to analyze and answer the following questions:#### Questions1. **What proportion of broilers weigh between 1150 and 1308 grams?**2. **What is the probability that a randomly selected broiler weighs more than 1510 grams?**3. **Is it unusual for a broiler to weigh more than 1610 grams?**#### InstructionsRound your answers to at least four decimal places.#### Part 1 of 3**Question:**What proportion of broilers weigh between 1150 and 1308 grams is __________.*Calculate the proportion of broilers using the normal distribution rules and enter your answer in the provided field.*#### Part 2 of 3**Question:**The probability that a randomly selected broiler weighs more than 1510 grams is __________.*Use the normal distribution formula to determine this probability and input your answer.*#### Part 3 of 3**Question:**Choose one (option) because the probability that a broiler weighs more than 1610 grams is __________.*Select the correct option based on your calculated probability and explain if this weight is unusual or not.***Options:**1. ( ) Usual because the probability is below a commonly accepted threshold.2. ( ) Unusual because the probability is above a commonly accepted threshold.**Note:** This content ensures you understand how to apply the normal distribution to real-world data, particularly in the context of agricultural science and commercial farming. The accompanying diagrams or graphs (if any) would visually represent the normal distribution curve and indicate the areas corresponding to the specific weights mentioned.**Save and Continue:**Click the "Continue" button after filling in all the answers and making the appropriate selection to submit your assignment.### Graph ExplanationIf a normal distribution curve is provided:1. **Mean and Standard Deviation:** - The center point of the curve represents the mean weight of 1387 grams. - The spread of the curve is determined by the standard deviation of 192 grams.2. **Shaded Regions:** - The region between 1150 and 1308 grams will be

Answered: Big chickens: The weights of broilers…

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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According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $109.54. Assume the amounts are normally distributed with standard deviation $25.00. Use the TI-84 Plus calculator to answer the following. (a) What proportion of bills are greater than $132? (b) What proportion of bills are between $90 and $145? (c) What is the probability that a randomly selected household had a monthly bill less than $129? Round the answers to at least four decimal places.
The u.s. Bureau of labor statistics reported the mean hourly wage rate for individuals in executive, administrative, and managerial occupations is $26. Assume the population mean is $26 and the population standard deviation is $4. A sample of 115 individuals in executive, administrative, and managerial occupation will be selected. What is the probability that the sample mean will be greater than $27?
Given a distribution with a mean of 80 and a standard deviation of 10, what is the probability of a score of 65 or less?
Almost all medical schools require applicants to take the Medical College Admission Test (MCAT). Scores on the MCAT are approximately Normal with a mean of 25 and a standard deviation of 6.5. If the School of Health Sciences interviews one WSSU student about their score on the MCAT. What is the probability that their score be between 21 and 30?
Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $108.87 . Assume the amounts are normally distributed with standard deviation $18.00 . Use the TI-84 Plus calculator to answer the following. (a) What proportion of bills are greater than $134 ? (b) What proportion of bills are between $83 and $145 ? (c) What is the probability that a randomly selected household had a monthly bill less than $118 ? Round the answers to at least four decimal places. Part 1 of 3 The proportion of bills that are greater than $134 is . Part 2 of 3 The proportion of bills that are between $83 and $145 is . Part 3 of 3 The probability that a randomly selected household had a monthly bill less than $118 is .
The sample standard deviation is a measure of what?
Please answer the attached question.
Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $108.87 . Assume the amounts are normally distributed with standard deviation $18.00 . Use the TI-84 Plus calculator to answer the following. (a) What proportion of bills are greater than $134 ? (b) What proportion of bills are between $83 and $145 ? (c) What is the probability that a randomly selected household had a monthly bill less than $118 ? Round the answers to at least four decimal places. Part 1 of 3 The proportion of bills that are greater than $134 is . Part 2 of 3 The proportion of bills that are between $83 and $145 is . Part 3 of 3 The probability that a randomly selected household had a monthly bill less than $118 is .
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?
You may need to use the appropriate appendix table or technology to answer this question. The weight of football players is normally distributed with a mean of 220 pounds and a standard deviation of 20 pounds. What is the probability of a player weighing less than 255 pounds
The NJ Department of Health has reported the average life span of NJ residents is 81 years. If the standard deviation is σ = 4 years, what is the probability of a person living to at least 92 years of age?
What is the standard deviation if the variance is 6.25? A, 2.5 B, 2.8 C. 3.5 D. 3.8

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