Homework Answers
The mean diameter of a ball bearing produced by a certain manufacturer is 0.80 cm with...
The mean diameter of a ball bearing produced by a certain manufacturer is 0.80 cm with a standard deviation of 0.03 cm. A sample of 36 ball bearings is randomly selected from a production run. What is the probability that the sample of ball bearings will have a mean of at most 0.815 cm? What is the probability that the sample of ball bearings will have a mean more than 0.81 cm?
Based on historical data, the diameter of a ball bearing is normally distributed with a mean...
Based on historical data, the diameter of a ball bearing is normally distributed with a mean of 0.527 cm and a standard deviation of 0.009 cm. Suppose that a sample of 36 ball bearings are randomly selected. Determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm. a. 0.9772 b. 0.0228 c. 0.5062 d. 0.0559
2. (3 points) The diameter of ball bearings produced in a manufacturing process at a particular...
2. (3 points) The diameter of ball bearings produced in a manufacturing process at a particular company can be explained using a uniform distribution. All the ball bearings produced have a diameter that ranges over the interval 3.5 milimeters to 5.5 milimeters. What is the probability that a randomly selected ball bearing will have a diameter greater than 4.4 milimeters? 3. (3.5 points) The monthly incomes of trainees at a local mill are normally distributed with a mean of $1050...
An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.73...
An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.73 inch. The lower and upper specification limits under which the ball bearing can operate are 0.72 inch (lower) and 0.74 inch (upper). Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.732 inch and a standard deviation of 0.006 inch. Suppose a random sample of 21 ball bearings are selected. Complete parts (a)...
1. A bearing used in an automotive application is supposed to have a nominal inside diameter...
1. A bearing used in an automotive application is supposed to have a nominal inside diameter of 3.81 cm. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 3.8037 cm. Bearing diameter is known to be normally distributed with standard deviation σ 0.03 cm. (a) Compute a 95%-confidence interval for the mean inside diameter. (b) Test the hypothesis H0 : μ = 3.81 versus H1 : μ关3.81 using α-001. What is the...
Is lack of sleep causing i want answer in ms excel 10:34 Done BUQU 1230 -...
Is lack of sleep causing
i
want answer in ms excel
10:34 Done BUQU 1230 - Assignment 2 F... Q1 Name Q2 Q3 Q4 Total Isrucr mat o of 25 a large company, iis asmed th 20% f the empleyees heild degees in b aple of 100pleyes is sclected, and thcir deprc ded A L The aple d th 18 mpyees hl degees in Gve the me its valae, and its poit estima a What is the popeiste p besines....
For Questions 1-4, let the random variable X follow a Normal distribution with mean u =...
For Questions 1-4, let the random variable X follow a Normal distribution with mean u = 200 variance 62 = 625. Q1. A random sample of n = 50 is obtained. What are the mean and variance of the sample mean, X-Bar? a. Mean ==> b. Variance ==> Q2. What is the probability that X-Bar is greater than 204? a. What is Z-Score for X-Bar greater than 204 ==> b. P[Z> Z-Score] ==> Q3. What is the probability that X-Bar...
2. (25 points) According to the manufacturer of M&M candies, a bag of M&Ms contains 465...
2. (25 points) According to the manufacturer of M&M candies, a bag of M&Ms contains 465 M&Ms with each M&M weighing 0.8535 grams. In order for the bag to contain the net total weight of 396.9 grams, the manufacturer must guarantee that each M&M's mean weight is at least 0.8535 grams. Assume the standard deviation is 0.0576. A random sample of 100 M&Ms were selected from a bag of 465 M&Ms and the mean value of the sample=0.8635. At the...
Statistic Q1(a) Mr. Sweetheart likes sugar in his hot tea. He buys sugar packets from a...
Statistic Q1(a) Mr. Sweetheart likes sugar in his hot tea. He buys sugar packets from a local grocery store. Suppose the amount of sugar in a packet follows a normal distribution with mean 2.17 grams and standard deviation 0.08 grams. The amount of sugar in a package should be between 2.13 and 2.21 grams. Otherwise, a packet is considered to be defective. i. What is the probability that a randomly selected packet is defective? ii. Three sugar packets are chosen...
The certain paper suggested that a normal distribution with mean 3,500 grams and standard deviation 590...
The certain paper suggested that a normal distribution with mean 3,500 grams and standard deviation 590 grams is a reasonable model for birth weights of babies born in Canada. (Use a table or technology.) (a) One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? (Round your answer to four decimal places.) (b) What is the probability...
2. (3 points) The diameter of ball bearings produced in a manufacturing process at a particular company can be explained using a uniform distribution. All the ball bearings produced have a diameter that ranges over the interval 3.5 milimeters to 5.5 milimeters. What is the probability that a randomly selected ball bearing will have a diameter greater than 4.4 milimeters? 3. (3.5 points) The monthly incomes of trainees at a local mill are normally distributed with a mean of $1050...
1. A bearing used in an automotive application is supposed to have a nominal inside diameter of 3.81 cm. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 3.8037 cm. Bearing diameter is known to be normally distributed with standard deviation σ 0.03 cm. (a) Compute a 95%-confidence interval for the mean inside diameter. (b) Test the hypothesis H0 : μ = 3.81 versus H1 : μ关3.81 using α-001. What is the...
Is lack of sleep causing
i
want answer in ms excel
10:34 Done BUQU 1230 - Assignment 2 F... Q1 Name Q2 Q3 Q4 Total Isrucr mat o of 25 a large company, iis asmed th 20% f the empleyees heild degees in b aple of 100pleyes is sclected, and thcir deprc ded A L The aple d th 18 mpyees hl degees in Gve the me its valae, and its poit estima a What is the popeiste p besines....
For Questions 1-4, let the random variable X follow a Normal distribution with mean u = 200 variance 62 = 625. Q1. A random sample of n = 50 is obtained. What are the mean and variance of the sample mean, X-Bar? a. Mean ==> b. Variance ==> Q2. What is the probability that X-Bar is greater than 204? a. What is Z-Score for X-Bar greater than 204 ==> b. P[Z> Z-Score] ==> Q3. What is the probability that X-Bar...
2. (25 points) According to the manufacturer of M&M candies, a bag of M&Ms contains 465 M&Ms with each M&M weighing 0.8535 grams. In order for the bag to contain the net total weight of 396.9 grams, the manufacturer must guarantee that each M&M's mean weight is at least 0.8535 grams. Assume the standard deviation is 0.0576. A random sample of 100 M&Ms were selected from a bag of 465 M&Ms and the mean value of the sample=0.8635. At the...
The certain paper suggested that a normal distribution with mean 3,500 grams and standard deviation 590 grams is a reasonable model for birth weights of babies born in Canada. (Use a table or technology.) (a) One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? (Round your answer to four decimal places.) (b) What is the probability...
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