population meAN = 7 P(X>6.42)= 0.591 FIND OUT THE VALUE OF STNADARD DEVATION FOR GIVEN NORMAL DISTRIBUTION
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population meAN = 7
P(X>6.42)= 0.591
FIND OUT THE VALUE OF STNADARD DEVATION FOR GIVEN
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- H0 : =150.00 kPa H1 :>150.00 kPa b) α= 0.01; sample variance : 1300 (freq. table ) sample deviation : 36.0555 (freq. table ) Calculation of Data Mean of the results= = 6700/50 =134 Mid Range of the results= (217.32+46.32)/2= 131.82 Range = (217.32-46.32) = 171 Modal Class= Class 121-145 with the highest frequency of 15 Median of the sorted set of data = 132.25 Mode = there are not clear modes In the samples. Sample variance S o= 1300 Sample deviation: S =36.55Tech Worx magazine reported that91%of software engineers rate the company they work for as "a good place to work." You are a headhunter, and as such you state that the percentage in the report is not correct. In a survey of220randomly selected software engineers, 207rated the company they work for as "a good place to work." Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the significance level of0.05, to support his claim that the ratio,p, of all software engineers who rate the company they work for as "a good place to work" is not91%. (to) Formulate the null hypothesish and the alternative hypothesish that you will use in your test. ho: 0 h₁:0 Р np= n61-p) = (c) carry out aZ -testand find thep-value. OA company that manufactures batteries used in electric cars is reporting that their newest model of battery has a mean lifetime, µ, of 18 years. To test the company's claim, a competitor has selected 56 of these batteries at random. The mean lifetime of the sample is 18.9 years. Suppose the population standard deviation of these lifetimes is known to be 4.8 years. Is there enough evidence to reject the claim that the mean lifetime of the newest model is 18 years? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis H, and the alternative hypothesis H,. Ho: 0 OA population of values a normal distribution with mu = 123.8 and Standard deviation = 23.8 A random sample of size n = 34 is drawn What is the standard deviation of the distribution of sample means Round to three decimalspopulation mean = 1.00 populaton standard deviation = 2.45 X = 1.43 N = 18 What is the probability (z score) for this value of X? .4325 .4286 .4247 .05Scores on the SAT exam approximate a normal distribution with a mean of 600 and a standard deviation of 80. Use the distribution to determine (z- score must be rounded to the two decimal places): The z-score for a SAT score of 470 The percent of SAT scores that fall above 750 The probability for a SAT score that fall below 800 The percentage of SAT scores that fall between 525 and 570The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 265 days and a standard deviation of 11 days. (a) What is the minimum pregnancy length that can be in the top 9%of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 3%of pregnancy lengths? (a) The minimum pregnancy length is ________ days.approximate area under the normal distribution curve that falls within one standard deviation above and below the mean. Please give the shaded region.Show complete solutionUsing Excel to calculate to 2 decimal places, if X follows a normal distribution with mean 55 and variance 16 and PDX > y) = 0.401294 then pic 5Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between -1.71 and -0.31 and draw a sketch of the region. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 1) Sketch the region. Choose the correct graph below. OA. -1.71 -0.31 The probability is (Round to four decimal places as needed.) OB. NEGATIVE Z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 02 .03 .04 .05 .06 .07 .08 .09 -1.71 -0.31 -3.50 and lower .0001 -3.4 0003 0003 -3.3 .0005 .0005 .0003 .0005 .0003 .0004 .0003 0004 .0003 .0003 .0003 .0003 .0002 .0004 .0004 .0004 .0004 .0003 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 -3.0 .0013 0013 .0013 0012 0012 .0011 .com .0011 0010 0010 -2.9 .0019 .0018 0018 .0017 0016 .0016 0015 .0015 0014…The test scores for the analytical writing section of a particular standardized test can be approximated by a normal distribution as shown in the figure (a) What is the maximum score that can be in the bottom 15% of scores ? (b) Between what two values does the middle 70% of scores lie?SEE MORE QUESTIONSRecommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON
A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON