to. They are being wasted. b. They need to be reprocessed. c. Acceptable.
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In the manufacture of steel rollers to fit an assembly, the external diameter is considered to be a random variable with mean equal to 23 mm and variance equal to 0.0036 cm2. 1.5 is supported standard deviations more than the mean and 3.5 standard deviations below average. A piece with a diameter below such specifications is considered waste, whereas a piece with a diameter larger than specified must be reprocessed. Which is the percentage of pieces that:
to. They are being wasted.
b. They need to be reprocessed.
c. Acceptable.
d. If you want to establish that the reprocessing is 5% of the rollers and waste of 1%, than specification limit would recommend.
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- A CNC (computer numerical control) machine produces iron automobile crankshafts. Samples are measured and the inner diameter are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. For all computations, assume an approximately normal distribution. The sample mean and standard deviation for the given data are = 1.0056, and s =0.0246. Find a 99% confidence interval on the mean of diameter. Compute a 99% prediction interval on a measured diameter of a single crankshaft piece taken from the machine. Find the 99% tolerance limits that will contain most of the metal pieces produced by the CNC machine. O a. CI on the mean: 0.9781 <µ <1.0331. Prediction Interval: 0.9186 s Xp+151.0926. Tolerance Interval (0.8937 and 1.1175) O b. Insufficient data to compute. Missing the sample size n Oc CIon the mean: 0.9781 <µ<1.0331. Prediction Interval: 0.9186 s Xp+151.0926. Tolerance Interval (0.8937 and 1.1175). Insufficient data to compute for the Tolerance Interval Od. CI on…A random sample of 16 statistics examinations from a large population was taken. The average score in the sample was 78.6 with a sample variance of 64. We are interested in determining whether the average grade of the population is significantly more than 75. Assume the distribution of the population of grades is normal . The value of the test statistic is A) 3.6 B) 1.8 C) 0.45The standard deviation of the distribution of sample means would be equal to:
- A random sample of 10 subjects have weights with a standard deviation of 11.9848 kg . What is the variance of their weights ? Be sure to include the appropriate units with the result .The average score on a statistics test was 2.52 with a standard deviation of 0.75. The teacher suspects that the exam was very difficult; Based on the above, he wants to adjust the scores so that the mean is 3.58 and the standard deviation is 0.25. Compute the values a and b such that the fit of type aX+b provide the desired mean and variance? Calculate: a=?b=?The mean tar content of a simple random sample of 25 unfiltered king-size cigarettes is 21.4 mg, with a standard deviation of 3 mg. The mean tar content of a simple random sample of 25 filtered 100-mm cigarettes is 13.0 mg with a standard deviation of 3.8 mg. The accompanying table shows the data. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Let population 1 be unfiltered king-size cigarettes. Complete parts (a) through (c) below. Click the icon to view the data. a. Use a 0.05 significance level to test the claim that unfiltered king-size cigarettes have a mean tar content greater than that of filtered 100-mm cigarettes. What does the result suggest about the effectiveness of cigarette filters? Identify the null and alternative hypotheses. O B. Ho: H1 =H2 O C. Ho: H1 #H2 H1: H1 =H2 O A. Ho: H1 = H2 H:H1 H2 O F. Ho: H1 = H2 H1:H1> H2 H:H1=H2 Hq: H1…
- Given a Normal Distribution with standard error of 8 from a population with a mean of 120 and a standard deviation of 48, then find the sample size (i.e. how many people or items were sampled from the population?).In an experiment the sorption was measured for a variety of hazardous organic chloroalkane solvents. The sample measurements were; 1.28, 1.35, 0.57, 1.16, 1.12, 0.91, 0.83, 0.43 Assuming that measurements are normally distributed. i. What is the mean of the sample for sorption of chloroalkane solvents. ii. What is the standard error of the mean for the sample for the organic chloroalkane solvents? iii. What the margin of error of the 90% confidence intervals for the population mean of sorption of chloroalkane solvents. iv. What is the upper limit for the 90% confidence intervals for the population mean of sorption of chloroalkane solvents.Given that the variances of random variables X and Y are 1600 and 900, respectively. The correlation between X and Y is 0.64. Find Var (X +Y).
- Companies who design furniture for elementary school classrooms produce a variety of sizes for children of different ages. Suppose the heights of kindergarten children are normally distributed with a mean of 97 cm and a standard deviation of 4.9 cm. What proportion of kindergarten children should the company expect to be less than 91 cm tall?Suppose the mean and variance of variable X are given respectively by μ = 30.9 and Var(X) =4.4. What is the average of the squared values of variable X?Given that the variances of random variables X and Y are 484 and 625, respectively. The correlation between X and Y is -0.52. Find Var (X-Y).
