life A sample of 100 dry battery cells tested to find the length he lollowing results X = 12 hours. = 3 hours Assuming the data to be normally distributed, what percentage of battery cells are expected to have life (1) More than 15 hours, (1) Less than 6 hours, and ( ) between 10 and 14 hours ? (Given z: 2.5 2 0.67 Area 0.4938 0.4772 0.3413 0.2487
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- A quality control officer compares the average lifetime of four different brandsof hand phones. For each of the brand, 5 hand phone was used to calculate theaverage lifetime in minutes. The results are given as follow V =756.55 W=3 X=16 Y=5.4 Z= 46.7006 i)Assuming the populations are normally distributed, determine whichaverage lifetime of the hand phones is significantly different at ? =0.05?A pharmcuticle company claims that its new drug reduces systolic blood pressure. The systolic blood pressure (in millimeters of Mercury) for 9 patients before taking the new drug and 2 hours after taking the drug are shown in the table below. is there enough evidence support the company's claim? Let D =(blood pressure before taking new drug)- (blood pressure after taking new drug).use significant levels of a=their 0.05 for the test. Assume that the systolic blood pressure levels are normally distributed for the population of patience both Before & After taking the new drug. 1.State the null and alternative hypothesis for the test. 2. Find the value of the standard deviation of the paired differences. Round to one decimal place. 3.compute the value of the test statistic. Round to three decimal places 4.determine the decision rule for rejecting the null hypothesis Ho. Round the numerical portion to three decimals. 5.make decision for the hypothesis test.Records for the last 15 years have shown that the average rainfall in a certain region of the country, for the month of March, to be 1.20 inches, with s = 0.45 inches. A second region had an average rainfall of 1.35 inches, with s = 0.54. estimate the difference of the true average rainfalls in those two regions as a 95% C.I. with the assumption of normal populations and unequal variances.
- A z-score of –1.6 has a corresponding Stanine score of:The article “Monte Carlo Simulation—Tool for Better Understanding of LRFD” (J. of Structural Engr., 1993:1586–1599) suggests that yield strength (ksi) for A36 grade steel is normally distributed with a mean of 50.2 ksi and a standard deviation of 3.8 ksi. What is the probability that the yield strength is greater than 60? (Use 4 decimal places) Show the area of interest in the density curve.An experiment to compare the spreading rates of five different brands of yellow interior latex paint available in a particular area used 4 gallons (J = 4) of each paint. The sample average spreading rates (ft2/gal) for the five brands were x1. = 462.0, x2. = 512.8, x3. = 427.5, x4. = 469.3, and x5. = 532.1. The computed value of F was found to be significant at level ? = 0.05. With MSE = 300.8, use Tukey's procedure to investigate significant differences between brands. (Round your answer to two decimal places.)
- For Normally distributed data with μ=194μ=194 and σ=0.8σ=0.8.20% of observations have values less than _____ .Incorrect Round to 1 decimal place.Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.For normally distributed data, what is the area greater than z = 0.34?The article “Monte Carlo Simulation—Tool for Better Understanding of LRFD” (J. of Structural Engr., 1993:1586–1599) suggests that yield strength (ksi) for A36 grade steel is normally distributed with a mean of 50.2 ksi and a standard deviation of 3.8 ksi. What is the probability that the yield strength is at most 45? (Use 4 decimal places). Show the area of interest in the density curve.
- I. Find the area corresponding to the following z values and sketch the curve. 1. P ( 0 < z < 1.47 ) 2. P( -2.35 < z < 0 ) 3. P( -2.59 < z < 0.41 ) 4. P ( -1.67 < z < 1.56 ) 5. P ( to the left of z = -1.96 ) II. Solve the problem carefully, sketch the curve. 1. The daily income of bank audition in SSS Commercial Bank is normally distributed with a mean of Php 675 and a standard deviation of Php 45. What is the z value for the income x of a bank auditor who earns Php 700 daily? 2. Johnny scored 72 in a quiz in Algebra for which the average score of the class was 65 with a standard deviation of 8. He also took a quiz in Statistics and scored 60 for which the average score of the class was 45, and the standard deviation was 12. Relative to other students in the class, did Johnny do better in Algebra or Statistics? PreviousNext) A random sample of 20 nominally measured of the length of a plastic ball is taken and the lengths are measured precisely. The measurements, in mare as follows 2.02 1.94 2.09 1.95 1.98 2.00 1.03 204 2,08 2,07 1.99 1.96 2.99 195 1.99 2.99 2.03 2.05 201 2.03 From the data, we caletilite Ex = 41.19, Ex* = 87.6977 Assuming that the lengths are normally distributed with unknown mean , and unknown variance 82 I find a 90% confidence interval for the variance of Interpret your results find a 90% confidence interval for the standard deviation Interpret your results.The diameter of a metal shaft used in a disk-drive unit is normally distributed with mean 0.2508 in. and std. dev. 0.0005 in. The specification on the shaft have been established as (0.25-0.0015 in, 0.25+0.0015in) If the process mean has been adjusted to 0.25 in, what fraction of the shafts produced does NOT conform to specifications? a). 0.0027 b). 0.0013 c).0.08 d). 0.16