# Probability, Sampling Distributions, and Inference

Feb 3rd, 2012
Studypool Tutor
Northcentral University
Course: a
Price: \$60 USD

Tutor description

Chapter Five 1) If light bulbs have lives that are normally distributed with a mean of 2500 hours and a standard deviation of 500 hours, what percentage of light bulbs have a life less than 2500 hours? 2) The lifetimes of light bulbs of a particular type are normally distributed with a mean of 370 hours and a standard deviation of 5 hours. What percentage of bulbs has lifetimes that lie within 1 standard deviation of the mean on either side? 3) The amount of Jen’s monthly phone bill is normally distributed with a mean of \$60 and a standard deviation of \$12. Fill in the blanks: 68% of her phone bills are between \$______________ and \$______________. 4) The amount of Jen’s monthly phone bill is normally distributed with a mean of \$50 and a standard deviation of \$10. Find the 25th percentile. 5) The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter g

Word Count: 1504
Showing Page: 1/9
Probability, Sampling Distributions, and InferenceChapter 51)The working of 1, 2 and 3 standard deviations (?) with a mean () of 2500 hours is given below and as graphically plotted there under.-1? to +1 ? is between 2000 hours and 3000 hours of light bulbs -2? to +2 ? is between 1500 hours and 3500 hours of light bulbs-3? to +3 ? is between 1000 hours and 4000 hours of light bulbsWe can observe from the above graph that the distribution is symmetric with a mean of 2500 hours, therefore, 50% of light bulbs will have a life of less than 2500 hours and the other 50% of light bulbs will have a life of more than 2500 hours.2)The standard deviation (?) is 5 hours and the mean () is 370 hours. We can solve this question by using the graph as above or it can be solved by using standard score (z) formula and its table. We will be using standard score (z) formula and table to solve the problem in order to get the precise percentage. The working of 1 standard deviation (?) with a mean () of 370 hours is as follows. -1? to +1 ? is between 365 hours (370 - 5) and 375hours (370 + 5) of light bulbsStandard score (z) = [(Data value - Mean) / Standard deviation]. Here we are going to find two standard scores i.e. 365 hours and 375 hours to find out the percentage on either side of 1 standard deviation. Standard score (z) of 365 hours will be [(365 - 370) / 5] = -1. Now we can refer the table of standard score (z) as given on page 211 of the course book (Bennett, Briggs & Tri

## Review from student

Studypool Student
" Outstanding Job!!!! "

1828 tutors are online

### Other Documents

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors