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Business Statistics: Mean, Standard Deviations

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I could use some help with the following. Could you please explain how you get the answer to the following questions in detail please so I can compare them to mine? I do not understand my book very well and I want to make sure I understand them correctly before I turn them in. Thank you

My first question is;

The mean starting salary for college graduates in the spring of 2004 was $36,280.
I am asking to assume the distribution of starting salaries follows the normal distribution with a standard deviation of $3,300.

What percentage of the graduates has starting salaries?

a) Between $35,000 and 45,000
b) More than $45,000
c) Between $40,000 and 45,000

My second question is;

I am ask to assume a binomial probability distribution with n=40 and (pi) =.55 and compute the following
a) The mean and standard deviation of 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

My third question;

A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fare charged for service from Hermosa Beach to the Los Angeles International Airport is $18.00 and the standard deviation is $3.50.

I am ask to select a sample of 15 fares and answer a and b. Please explain how you got your answer so I can compare it to mine.

a) What is the likelihood that the sample mean is btw $17.00 and 20.00?
b) What must you assume to make the above calculation?

My last question;
Dr. Patton is a Professor of English. Recently he counted the number of misspelled words in a group of student's essays. For his class of 40 students, the mean number of misspelled words was 6.05 and the standard deviation 2.44 per essay.

I need help constructing a 95 percent confidence interval for the mean number of misspelled words in the population of student essays.

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Solution Summary

This solution provides a thorough walk-through of the listed problems.

Solution provided by:
Changping Wang, MA
Education
  • 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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