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Finding 95% confidence interval using normal distribution.
At a large university, students have an average credit card debt of $2500, with a standard deviation of $1200. A random sample of students is selected and interviewed about their credit card debt. Use the Rule to answer the question about the mean credit card debt for the students in this sample.
If we imagine all the possible random samples of 250 students at this university, 95% of the samples should have means between what two numbers?
o $2348.22 and $2651.78o $300 and $4900o $250.00 and $2651.78o $250.00 and $2575.89o $2272.33 and $2727.67
Find the probability that a women develops breast cancer given that she has a family history of breast cancer._____________
Determine the special property of standard normal distribution, compared to other normal distributions?
Find out the range (R), variance (V) and standard deviation (s) for the given set of values:
What is the expected shape of the distribution of the sample mean?
A bag of jelly belly candies contains the following colored jelly beans: red (10), blue (2), orange (5), brown (21), green (0), and yellow (18). Construct the probability distribution for x.
In addition, record the values for the UCL, center line, and LCL. Also, record any points that are out of control. Use a z-value of three.
Determine the range and standard deviation of the times.
The calculated Z test statistic is a positive value that leads to a p-value of .045 for the test. If the significance level (α) is .01, the null hypothesis would be rejected.
Use a 95% confidence interval for multiple comparisons. Show your calculation by hand between Monday and Tuesday. Match your answer with Minitab output. Attach Minitab output.
Iit is known that signal change in this brain area is normally distributed with mean of 35 and standard deviation of 10. By using .01 level, what must researcher conclude?
What is the p-value associated with the above sample result?
Find out the variance for percentage of tickets? Determine the probability that your sample will have the average between 70 and 75?
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