A population of values has a normal distribution with u= 95.8 and o = 21.3. You intend to draw a randomsample of size n = 106.Find the probability that a single randomly selected value is between 92.1 and 100.1.P(92.1 « X < 100.1) =Find the probabitity that a sample of stze n= 106 is randomly selected with a mean between 92.1 and 100.1.P(92.1 < M< 100.1) -Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or zscores rounded to 3 decimal places are accepted.

Answered: Image /qna-images/answer/b883f92e-5e8f-4bc5-9572-7ba0964ad57c.jpg

Answered: A population of values has a normal…

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 company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 269.2-cm and a standard deviation of 0.7-cm. For shipment, 17 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is less than 269.4-cm.P(M < 269.4-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
A population of values has a normal distribution with �=105.6 and �=10.5. Find the probability that a single randomly selected value is greater than 102.1. Round your answer to four decimal places. �(�>102.1)= Find the probability that a randomly selected sample of size �=66 has a mean greater than 102.1. Round your answer to four decimal places. �(�>102.1)=
A population of values has a normal distribution with �=207.6 and �=10.3. You intend to draw a random sample of size �=62.Find the probability that a single randomly selected value is greater than 209.4.P(�< 209.4) = Find the probability that a sample of size �=62 is randomly selected with a mean greater than 209.4.P(�¯< 209.4) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.7-cm and a standard deviation of 1.4-cm. For shipment, 7 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is between 111.5-cm and 112-cm.P(111.5-cm < M < 112-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
CNNBC recently reported that the mean annual cost of auto insurance is 1013 dollars. Assume the standard deviation is 230 dollars. You take a simple random sample of 62 auto insurance policies. Find the probability that a single randomly selected value is less than 998 dollars. P(X Next Question
Q11
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 254.8-cm and a standard deviation of 1.3-cm. For shipment, 8 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 253.9-cm and 254-cm. P(253.9-cm < M < 254-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Suppose that the hibernation time for black bears is normally distributed with mean 150 days and standard deviation 30 days. 12 randomly selected black bears are monitored. Round all answers to two decimal places. A. ~ N( 1) B. For the 12 black bears, find the probability that the average hibernation time is more than 170 days. C. What is the probability that one randomly selected black bear will hibernate longer than 170 days? D. Find the IQR for the average of 12 bears. Q1 = Q3 = IQR:
For patients with a certain disease the amount of certain protein in the blood is normally distributed with a mean of 130 and a standard deviation of 5. For patients without the disease the amount of protein that is normally distributed with a mean of 115 and a standard deviation of 4. Where would you want the "Tested Positive" to start so that the probability of a person with the disease will not test positive is 0.0055. Answer to 3-decima places.
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 163.9-cm and a standard deviation of 1.6-cm. For shipment, 20 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is between 163.3-cm and 165-cm.PP(163.3-cm < ¯xx¯ < 165-cm) = Enter your answer as a number accurate to 4 decimal places.
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 266.3-cm and a standard deviation of 1.7-cm. For shipment, 22 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is less than 265.8-cm.P(M < 265.8-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 269.6-cm and a standard deviation of 2.1-cm. For shipment, 8 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 267.9-cm and 272-cm. P(267.9-cm < M<272-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Ouestion Lloln

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