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Find these probabilities using standard normal distribution.
The average credit card debt for college seniors is $3262. If the debt is normally distributed with a standard deviation of $1100, find these probabilities:
a. That the senior owes at least $1000.b. That the senior owes more than $4000.c. That the senior owes between $3000 and $4000.
We are interested in comparing the proportions of males and females who think earning a large salary is very important to them. I surveyed 200 of each gender and recorded their answers to the question as yes or no.
Test the hypothesis that no difference exists between the groups at α = .05 and perform and report all related tests and statistics.
Assume that the smiling time of eight-week old babies, in seconds, follows a uniform distribution between 0 and 23 seconds, inclusive. This means that any smiling time from 0 to and including 23 seconds is equally likely.
Probability using discrete distribution.The probability distribution for the random variable x follows.
Develop a joint probability table for these data. If a student goes full time, determine the probability that school quality is first reason for selecting a school?
How large random sample of potential voters will be required to evaluate result with 95% confidence and with margin of error of plus and minus 3%?
At the .01 significance level, is there a difference in the use of the four entrances?
A sample of 64 Atlanta consumers showed a mean quarterly water bill of $51 with a sample standard deviation of $12. At α = .05, does the Atlanta sample support the conclusion that above-average rates exist for this private water system?
Find out the range (R), variance (V) and standard deviation (s) for the given set of values:
There were 3 defectives in the first batch and 5 in the second. Find a 95% confidence interval for the difference in the proportion of defectives.
Is there evidence of violations of the usual ANOVA assumptions of equal variances and normal populations? Set up and perform appropriate TESTS at the α = 0.05 level of significance.
Assume that the number of left-handers is a binomial random variable. (Refer to the formula for mean and standard deviation for binomial prob. Distribution)
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