Known: Variable X: Mean = 6, Standard Deviation = 0,6 and Normal Distribution Variable Y: Mean = 3, Standard Deviation = 0,3 and Lognormal Distribution
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Known:
Variable X:
Variable Y:
Mean = 3, Standard Deviation = 0,3 and Lognormal Distribution
Find:
a. The probabilty of Y will exceed 6
b. The probabilty of X will exceed 8
c. Calculate X and Y 90th percentile value
d. Calculate X and Y 50 year return period
Step by step
Solved in 2 steps
- Draw a normal curve with a mean of 66 and a standard deviation of 13. Describe how you constructed the curve and discuss its features The normal distribution curve is centered at ( ? ) and has 2 points of inflection,( ? ) representing mu minus sigma, and ( ? ) representing mu plus sigma.X is the number of red marbles that Suzan has in her hand after she selects 5 marbles from a bag containing 5 red marbles and 2 green ones. expected value variance standard deviationGrowth of bacteria culture after one week, represented by its size, is known to follow lognormal distribution with mean of 2 and variance of 0.2. What is the upper and lower quartile of the size of the culture after one week?
- H0 , test statistic, and p value pleaseTimes for a procedure are normally distributed. There are 2 methods. Method A has a mean of 25 minutes and a standard deviation of 8 minutes. Method B has a mean of 29 minutes and a standard deviation of 4 minutes. Which procedure is preferred if it must be completed within 33 minutes?The one in the yellow is the previous tab.
- Scores on a history test have an average of = 70 with standard deviation of 5. What is the z-score for a student who earned a 65 on the test? What is the coefficient of variation? 2. The annual salaries of employees in a large company are approximately normally distributed with a mean of $70,000 and a standard deviation of $30,000. (6 points) What percent of people earn less than $40,000? What percent of people earn more than $85,000? What percent of people earn between $40,000 and $85,000? 3. Assume women’s heights follow a normal distribution with a mean of u = 64.5 inches and a standard deviation of 2.5 inches. Find the probability that a sample of 14 women have a mean height greater than 65 inches.A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage x¯1x¯1 = $ 1,101 x¯2x¯2 = $ 1,766 Sample Standard Deviation s1 = $ 696 s2 = $ 838 Repair Incidents n1 = 12 n2 = 9 (a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round answers to 3 decimal places. Negative values should be indicated by a minus sign.) (b) Repeat part (a), using the assumption of unequal variances with Welch's formula for d.f. (Round answers to 2 decimal places. Negative values should be indicated by a minus sign.) (d) Construct separate…es point(s) possible Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 22 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.14 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.98 oz and 12.70 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.18 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.18 oz. Use a 0.05 significance level. Complete parts (a) through (d) below. b. Compute the test statistic. x²=0 (Round to three decimal places as needed.) c. Find the P-value. P-value= (Round to four decimal places as needed.) d. State the conclusion. the level of significance. There is evidence to conclude that the population Ho, because the P-value is standard deviation of…
