Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
2. Probability Distributions and Frequency Tables
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Exercise 18 Page 834

Practice makes perfect
a We are given a table showing the results of a survey of 80 households in Westville. First we want to find the probability distribution of the number of computers in Westville households. To do so we need to calculate the relative frequencies for all numbers of computers. Let's start with rearranging our table a bit.
Computer Survey
Number of Computers 0 1 2 3 More than 3
Frequency 12 29 31 6 2

To find the relative frequencies we will divide the frequency for the particular number of computers by the total of the frequencies. Since 80 households took part in the survey, the total of frequencies is 80. The relative frequencies represents a probability of picking a household with a certain number of computers.

Computer Survey
Number of Computers 0 1 2 3 More than 3
Frequency 12 29 31 6 2
Probability 12/80=3/20 29/80 31/80 6/80=3/40 2/80=1/40
b Now we want to predict the number of households that will have exactly three computers if Westville has 15 200 households. Note that in the previous part of exercise we calculated an experimental probability for this event. It is experimental because we based it on the actual results of the survey. Consider the obtained table.
Computer Survey
Number of Computers 0 1 2 3 More than 3
Frequency 12 29 31 6 2
Probability 3/20 29/80 31/80 3/40 1/40
The experimental probability of selecting a household with exactly three computers is 340. To predict the number of households with exactly three computers we need to multiply this probability by the number of households in Westville, 15 200.
Households with3 computers=P(3computers) * Total number of households
Households with3 computers= 3/40 * 15 200
Households with3 computers=45 600/40
Households with3 computers=1140
The predicted number of households in Westville with exactly 3 computers is 1140.
c Finally we want to predict the number of households with either two or three computers. We will do it analogically as in the previous part of the exercise. Consider the obtained table.
Computer Survey
Number of Computers 0 1 2 3 More than 3
Frequency 12 29 31 6 2
Probability 3/20 29/80 31/80 3/40 1/40
This time the probability of picking a household with either two or three computers will be the sum of probability of picking a household with two computers, 3180, and the probability picking a household with three computers, 340. Let's calculate this sum.
P(2or3computers)=P(2computers)+P(3computers)
P(2or3computers)= 31/80+ 3/40
P(2or3computers)=31/80+6/80
P(2or3computers)= 37/80
Again, this is an experimental probability because it is based on the results of the survey. To predict the number of households with either two or three computers we need to multiply this probability by the total number of households in Westville, 15 200.
Households with2or3 computers=P(2or3computers) * Total number of households
Households with2or3 computers= 37/80 * 15 200
Households with2or3 computers=562 400/80
Households with2or3 computers=7030
The predicted number of households in Westville with either 2 or 3 computers is 7030.