Avril and Ben are surveying passing traffic, noting the number of occupants in each vehicle. Below is the table of results of their survey?

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Avril and Ben are surveying passing traffic, noting the number of occupants in each vehicle. Below is the table of results of their survey?

Avril and Ben are surveying passing traffic, noting the number of occupants in each vehicle. Below is the table of results of their survey.

(i) How many vehicles passed Avril and Ben?

Based on these results, what is the probability that the next vehicle passing by:

(ii) has 2 occupants?

(iii) has 4 occupants?

(iv) has less than 4 occupants?

(v) has at least 3 occupants?

vi) If 6 000 vehicles pass the same spot, how many would one expect to contain more than 3 occupants? Answer to the nearest 10 vehicles.

vii) Avril did some calculations and expects that on Tuesday there will be 3520 vehicles without passengers pass the same spot. (ie. vehicles with only the driver in them). How many vehicles in total does Avril expect to pass on Tuesday?

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Answer 1 · 2017-01-02T02:39:55

This is all about expectation - the chart has given us a sample and now we can use it to predict outcomes. See below for details:

Explanation:

This is all about expectation - the chart has given us a sample and now we can use it to predict outcomes.

i - We need to know the total number of cars that are in the sample so that we can use that for the ratios and fractions in the following questions:

$44 + 37 + 22 + 10 + 6 + 1 + 4 = 124$ $44+37+22+10+6+1+4=124$

ii: has 2 occupants: $\frac{37}{124}$ $37/124$

iii: has four occupants: $\frac{10}{124} = \frac{5}{62}$ $10/124=5/62$

iv: has less than 4 - and so it's 1, 2, and 3 occupants: $\frac{105}{124}$ $105/124$

v: has at least 3 occupants - and so it's 3, 4, 5, 6, and more: $\frac{43}{124}$ $43/124$

vi: if 6,000 cars pass by, how many would be expected to have more than 3 occupants?

With our sample, we have $\frac{21}{124}$ $21/124$ that have more than 3 occupants. So we'd expect (using a ratio of our sample $\frac{21}{124}$ $21/124$ and comparing it to a ratio of what we'd expect to the total number of cars expected):

$\frac{21}{124} = \frac{x}{6000}$ $21/124=x/6000$

$124 x = 21 \times 6000$ $124x=21xx6000$

$x = \frac{21 \times 6000}{124} ≅ 1016 ≅ 1020$ $x=(21xx6000)/124~=1016~=1020$ rounded to the nearest 10

vii: if 3520 single passenger vehicles are anticipated on a single day, the total number of cars anticipated is (using a ratio of single passenger cars in our sample and comparing it to a ratio of anticipated single passenger cars to the total number of cars):

$\frac{44}{124} = \frac{3520}{x}$ $44/124=3520/x$

$44 x = 124 \times 3520$ $44x=124xx3520$

$x = \frac{124 \times 3520}{44} = 9920$ $x=(124xx3520)/44=9920$

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Avril and Ben are surveying passing traffic, noting the number of occupants in each vehicle. Below is the table of results of their survey?

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