Week 3 Assignment: Apply Probability, Sampling Distributions, and Inference

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Probability and sampling distribution Name Institutional affiliations Chapter Five Quiz 1 Solution. Mean=2500 Standard deviation=500 By using the 68-95-99.7 rule, nearly 68% of the light bulbs have a life time between: Mean- standard deviation=2500-500 =2000 Mean+ standard deviation=2500+500 =3000 An approximate number of 95% have a life of between Mean-2 standard deviations =2500-(2*500) = 1500 and Mean+2 standard deviations=2500+ (2*500) =3500 95.7% of the bulbs have a life of between Mean-3 standard deviations=2500-(3*500) =1000 and Mean +3 standard deviations=2500+3 standard deviations =2500+ (3*500) =4000 This means that 49.85% (95.7%/2) of the bulbs have a life that is inferior to 2500 (mean) Z-score = value – (Mean / SD) =2500-2500 500 =0 From the z table 50% is the corresponding percentile of bulbs with life of less than 2500 hours. Quiz 2 Solution Mean = 370 Standard deviation=5, Standard deviation = 1 68-95-99.7 rule states that: 68% of all the observations lie within 1 SD of the mean. =370-5 =365 hours AND =370+5 =375 hours. Z (for 365 value) = (365-370)/5 =-1 which corresponds to 15.87% percentile of the Z score table Z (for the 375 value) = (375-370)/5 =1 which corresponds to 84.13% percentile of the Z score table Therefore: 84.13%-15.87% = 68.26% of all scores fall between 1 standard deviation of the mean on either side. Question 3 Solution: Mean=\$60 Standard deviation= \$12 60-12 = 48 and 60+12 = 72 \$48 and \$72 Quiz 4 Solution Mean = \$50, Standard deviation = \$10 Data value = \$50/\$10 25th percentile z-score =-0.674 -0.674= (y - 50) / 10 6.74 =

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