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Cumulative Binomial Probability Distributions

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A manufacturer of window frames know from long experience that 5% of the production will have some type of minor defect that will require an adjustment. What is the probability that in the sample of 20 window frames:

A. None will need adjustment?
B. At least one will need adjustment?
C. More than two will need adjustment?
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Hypergeometric Probabilities

A population consists of fifteen items, ten of which are acceptable. In a sample of 4 items, what is the probability that exactly 3 are acceptable? Assume the samples are drawn without replacement.
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The mean starting salary for college graduates in the spring of 2004 was 36,280. assume that the distribution of starting salaries fallows the normal distribution with a standard deviation of 3,300. What percent of the graduates have starting salaries:
A. Between 35,000 and 40,000?
B. More than 45,000?
C. Between 40,000 and 25,000?
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Assume a binomial probability distribution with n=40 and P=.55. Compute the following:

A. The mean and standard deviation of the random variable.
B. The probability that X is 25 or greater.
C. The probability that X is 15 or less.
D. The probability that X is between 15 and 25 inclusive.

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A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fair charged for service from Hermosa Beach to the Los Angeles International Airport is $18.00 and the standard deviation is $3.50. We select a sample of 15 fairs.

A. What is the likelihood that the sample mean is between $17.00 and $20.00?
B. What must you assume to make the above calculation?

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Dr. Patton is Professor of English. Recently he counted the number of misspelled words in a group of student essays. For his class of 40 students, the mean number of misspelled words was 6.05 and the standard deviation 2.44 per essay. Construct a 95 percent confidence interval for the mean number of misspelled words in the population of student essays.

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Solution Summary

The solution solves a few questions concerning cumulative binomial probability distributions as well as hypergeometric probabilities

Solution provided by:
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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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