Question 6 < > For a standard normal distribution, find: P(z > -1.29) Express the probability as a decimal rounded to 4 decimal places.
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Q: standard normal distribution use excel
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- Standard normal distribution, to find the area of the shaded region, the graph shows the standard normal distribution of bone density scores with mean 0 and standard deviation 1 z= 1.07 and z=0.67True or false? How?Standard Normal Distribution Using the standard normal distribution, find:P(z > 1.87) Enter answer to 4 decimal places.
- The test statistic of z = -2.21 is obtained when testing the claim that p<. This is a left-tailed test. Using a 0.01 significance level, complete parts (a) and (b). Click here to view the standard normal distribution table for negative z scores. Click here to view the standard normal distribution table for positive z scores. a. Find the critical value(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) OA. There are two critical values; the lower critical value is and the upper critical value is B. There is one critical value; the critical value isCOURSE LEARNING OUTCOME Perform a hypothesis test for two population proportions. Give the null hypothesis, alternative hypothesis, reject or fail to reject the null hypothesis, and give the conclusion. a) For the Claim P1< P2 state the null hypothesis and alternative hypothesis in symbols b) For the P-value = 0.073, state whether you reject or fail to reject the null hypothesis if the significance level is 0.05 and why. c) State the conclusion of the problem.Assume 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…
- A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 165 randomly selected fans on their way out of the event and asks them to rate their satisfaction with the event on a scale from 0 to 10. The average satisfaction rating is 6.9. It is known from previous studies of this type that the standard deviation in satisfaction level is 2.5. Calculate the margin of error and construct the 95% confidence interval for the true mean satisfaction level for the sporting event. Standard Normal Distribution Table ΕΞ Dound in th SAMSUNG SUBMIT QUESTION THES SAVE PROGRESS SUBMITThe chickens at Colonel Thompson's Ranch have a mean weight of 1950 g, with a standard deviation of 150 g. The weights of the chickens are closely approximated by a normal curve. Find the percent of all chickens having weights more than 1815 g. Click here to view page 1 of the standard normal table. Click here to view page 2 of the standard normal table. XX It is expected that percent of all chickens have weights of more than 1815 g. (Type an integer or decimal rounded to two decimal places as needed.)Significance For bone density scores that arc normally distributed with a mean of 0 and a standard deviation of I, find the percentage of scores that are a. significantly high (or at least 2 standard deviations above the mean). b. significantly low (or at least 2 standard deviations below the mean). c. not significant (or less than 2 standard deviations away from the mean).