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14 October, 04:30

4.30 Support for casino in Toronto. In an effort

to seek the public's input on the establishment of a

casino, Toronto's city council enlisted an independent

analytics research company to conduct a public survey.

A random sample of 902 adult Toronto residents were

asked if they support the casino in Toronto. Here are

the results:

Response

Strongly

support

0.16

Somewhat

support

0.26

Mixed

feelings

Probability

?

Response

Somewhat

oppose

0.14

Strongly

oppose

0.36

Don't

know

0.01

Probability

(a) What probability should replace "?" in the distribution?

(b) What is the probability that a randomly chosen adult

Toronto resident supports (strongly or somewhat) a casino?

+3
Answers (1)
  1. 14 October, 07:04
    0
    a) 7% or 0.07

    b) 42% or 0.42

    Explanation:

    Let's begin by listing out the given information:

    Population = 902,

    Probability distribution

    Strongly support = 0.16 or 16%,

    Somewhat support = 0.26 or 26%,

    Mixed feelings = ?,

    Somewhat oppose = 0.14 or 14%,

    Strongly oppose = 0.36 or 36%

    Don't know = 0.01 or 1%

    a) The summation of every probability of an event (survey) is 100% ⇒ Pr (survey) = 100%

    Pr (survey) = Pr [strongly support + somewhat support + mixed feelings + somewhat oppose + strongly oppose + don't know]

    100 = 16 + 26 + x + 14 + 36 + 1

    To obtain the probability of the 'Mixed feelings' populace, we subtract the summation of all other probabilities from 1 or 100%

    x = 100 - (16 + 26 + 14 + 36 + 1) = 100 - 93

    x = 7% or 0.07

    ∴ the probability of the 'Mixed feelings' populace is 7% or 0.07

    b) Probability of a random adult supporting is given by the summation of the probabilities of the adults who strongly support & those who somewhat support

    Pr (support) = Pr (strongly support) + Pr (somewhat support)

    Pr (support) = 0.16 + 0.26

    Pr (support) = 0.42 or 42%

    ∴ the probability that a randomly chosen adult supports the casino in Toronto is 42%
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