1. The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 29.75 inches long? (Points : 5) .8944 .1056 .9332 .0668 | 2. The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.
Practice Midterm Exam Page 1 1. | Indirect labor is a part of: | A | Prime cost. | B | Conversion cost. | C | Period cost. | D | Nonmanufacturing cost.
Yes, it is significant because as indicated by the asterisk, p <0.05 is the least acceptable value for statistical significance. 3. Which t ratio listed in Table 3 represents the smallest relative difference between the pretest and 3 months? Is this t ratio statistically significant? What does this result mean?
$48.00 | SELLING PRICE PER UNIT | $28.49 | VARIABLE COST PER UNIT | $19.51 | UNIT CONTRIBUTION MARGIN | Finally to determine the breakeven point you divide the Total Fixed cost by the Unit Contribution Margin. $260,000.00 | TOTAL FIXED COST | $19.51 | UNIT CONTRIBUTION MARGIN | 13,326 | UNITS TO BREAKEVEN | CONCLUSION In order for Waltham Motors to breakeven they must manufacture a minimum of 13,326 units to cover all cost. Question 2 * Using budget data, what was the total expected cost per unit if all manufacturing and
The random sample of 65 satisfaction rating yields a sample mean of x = 42.954. Assuming that S = 2.64, use critical values to test H0 versus Ha at each of a = 10, .05, .01 and .00l. MU = 42, N = 65, X-bar = 42.954, sigma = 2.64 Z= (x-bar – mu)/(sigma/sqrt n) Z = (42.954 – 42)/(2.64/sqrt65 = 2.9134. Wk4/Assignment 9.13 continued Critical upper tail = Z – scores for 1.2816, 1.6449, 2.3263 and 3.092 for a = 0.10, 0.05, 0.01 and 0.001. Since 2.9134>1.2816, 1.6449 and 2.3263, I rejected H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42.
The confidence interval for the first group mean is thus (4.1, 13.9). Similarly for the second group, the confidence interval for the mean is (12.1, 21.9). Notice that the two intervals overlap. However, the t-statistic for comparing two means is: t= 17 − 9 2.5 2 + 2.5 2 = 2.26 which reflects that the null hypothesis, that the means of the two groups are the same, should be rejected at the α = 0.05 level. To verify the above conclusion, consider the 95 percent confidence interval for the difference between the two group means: (17 − 9 ) ± 1.96 × 2.5 2 + 2.5 2 which yields (1.09, 14.91).
Assignment #1: JET Copies Case Problem MAT 540 Quantitative Methods Professor Abdul Khan January 26, 2014 1. In Excel, use a suitable method for generating the number of days needed to repair the copier, when it is out of service, according to the discrete distribution shown. As for the first segment of the case, I needed to figure out the number of days needed in order to repair the copier. Initially, I assumed that the amount of days required to repair a copier is random. So with that, I could generate a random number using the Excel RAND function that I designated r2 between 0 and 1.
Find the mean and the standard deviation of the sampling distribution of the sample mean of bar x. Mean = 20 (given in the question) Standard deviation = 4/√64 Standard deviation = 4/ 8 Standard deviation = 0.5 c. Calculate the probability that we will obtain a sample mean greater than 21; that is, calculate P(x > 21). Hint: Find the z value corresponding to 21 by using μ and σ because we wish to calculate a probability about x. Then sketch the sampling distribution and the probability Z = (21-20)/ (4/√64) Z = 1/0.5 Z = 2 P (z>2) P = 1-0.9772 P = 0.0228 d. Calculate the probability that we will obtain a sample mean less than 19.385 ; that is calculate P(x.< 19.385). Z = (19.385 - 20)/ (4/√64) Z = -0.615 / 0.5 Z = -1.23 P = 0.1093 7.30 On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympic Games in Salt Lake City, Utah.
BMI=25.96 this is what I came up with as my BMI figure in the Kg/Meters. “One Kilogram equal 2.2 Lbs and 1 inch equal .0254 meters”. To convert your height by to inch 1 foot =12 inches: 5 foot *12 inches = 60 inches that is how you come up with the inches. Pounds to Kilograms: 132.50 lbs/2.2 =60.227 kgs which basically means 132.50 is equivalent to 60.227 kilograms. To figure out if your over weight do the formula just like I did above and if you are: •
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: 0.2 is equal to 1/5 or 20%, 0.2 indicates 20% of the data values are in this particular interval. 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.