Probability, Sampling Distributions, and Inference

Feb 3rd, 2012
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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

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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

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