#8,9,10 Changer 5 introduce the concept of probability distribution with a focus
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#8,9,10 Changer 5 introduce the concept of probability distribution with a focus on decrease probability distributions. The relationship between random variables and probability distribution is a key from this chapter, probability distribution and the distribution are two types of decrease probability distribution that are described. The table below lists the number of vehicles per house hold in the state of California and the united states overall. (The columns representing 4 cars actually comes from data for "4 or more" cars. But we will consider that column to represents 4 cars for the purpose of these exercises) The members in the now for are numbers of households, so in there are 18, 807 households with no cars, 70, 144 households with one car, and so on. Assume that one of the 171, 361 household in is randomly selected. Find the probabilities that the household has 0 cars, 1 car, 2 cars, 3 cars, 4 cars. Use the results to construct a table representing the probability distribution for sacraments. Find the mean and standard deviation of the probability distribution. Using the results form part (b) with the range rule of thumb, is it unusual to randomly select a household in and obtain one with 3 cars Explain. Using the results from part (b) with the range rule of thumb, is It unusual to randomly select a household in and obtain one with 4 cars Explain. Repeat Exercise I using California instead of Sacramento Repeat Exercise 1 using the united instead of compare the probability distributions from exercise 1, 2 and 3, any differences? Using the data for Sacramento in the above table, find the probability then a randomly selected household in Sacramento does not have a car, Round the rules to three decimal places. Using the data for Sacramento in the above table, find the probability that a randomly selected household in Sacramento has one or more cars. Rounded the rules to three decimal places. Using the rounded rules from parts (a) and (b), find the probability that among 5 randomly selected households in Sacramento exactly three have one or more cars. Repeat Exercise 5 using California instead of Sacramento Repeat Exercise 5 using the United states increased of Sacramento Next consider the mode of transportation that employees age 16 and older me to get to work. The table below summaries the data for Los Angeles, California and the United States. Find the probability that a randomly selected employee from Loss Angeles use public transportation. Round the result to three decimal places. Find the probability that a randomly selected employee from loss Angeles does not use public transportation. Round the results three decimal places. Using the rounded results from parts (a) and (b). find the probability that if you randomly select 10 from los Angeles, exactly two of them use public transportation. Using the rounded results from parts (a) and (b) find the probability that if you randomly select 10 commuters from Los Angeles, at least one of them uses public transportation. Repeat Exercise 8 using California involved of los Angeles. Repeat Exercise 8 using the United State instead of loss Angeles.Explanation / Answer
8.
a. Prob that person from LA chooses public transport = 160809/(160809+1243247) = 0.114
b. Prob = 1-0.114 = 0.886
c. Out of 10, 2 use public transport and 8 do not .
Hence prob = 10C2*0.144^2*0.886^8 = 0.3543
d. Out of 10, at least 1 uses public transport = 1-Prob that none uses public transport = 1 - 10C10*0.886^10 = 0.702
9.
a. Prob that person from California chooses public transport = 777658/(777658+13776039) = 0.053
b. Prob = 1-0.053 = 0.947
c. Out of 10, 2 use public transport and 8 do not .
Hence prob = 10C2*0.053^2*0.947^8 = 0.0818
d. Out of 10, at least 1 uses public transport = 1-Prob that none uses public transport = 1 - 10C10*0.947^10 = 0.4199
10.
a. Prob that person from USA chooses public transport = 6306526/(6306526+125034525) = 0.048
b. Prob = 1-0.048 = 0.952
c. Out of 10, 2 use public transport and 8 do not .
Hence prob = 10C2*0.048^2*0.952^8 = 0.0699
d. Out of 10, at least 1 uses public transport = 1-Prob that none uses public transport = 1 - 10C10*0.952^10 = 0.3885
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