Question

A survey of 1,084 adults was asked, "Do you feel overloaded with too much information?" Suppose that the results indicated that of 540 males, 240 answered yes. Of 544 females, 276 answered yes. Construct a contingency table to evaluate the probabilities. Complete parts (a) through (d). a. What is the probability that a respondent chosen at random indicates that he/she feels overloaded with too much information? (Round to three decimal places as needed.) b. What is the probability that a respondent chosen at random is a female and indicates that he/she feels overloaded with too much information? (Round to three decimal places as needed.) c. What is the probability that a respondent chosen at random is a female or who feels overloaded with too much information? (Round to three decimal places as needed.) d. What is the probability that a respondent chosen at random is a male or a female? (Round to three decimal places as needed.)

Answer 1

First we prepare the contingency table that just shows us all the details we have gathered :

	Said Yes	Said No	Total
Male	240	300	540
Female	276	268	544
Total	516	568	1084

Now Probability of any event can simply be calculated by :

  $Probability \,\,of \,\,Event\, A = \frac{No.\,\,of \,\,favourable \,\,outcomes \,\,for\,\,event\,\,A}{Total\,\,no.\,\,of \,\,outcomes}$

(a) We need to calculate probability that a respondent chosen at random said Yes to the question .

   No , of people who said yes = 516

Probability of a person chosen ,having said yes =

516 No. of people who said yes Total no. of people 0.476 1084

(b) We need probability that a person chosen at random is a female who said yes to the question .

   No , of Females who said yes to the question = 276

   Probability of a person chosen , being female and said yes

No. of females who said yes Total no. of people

$=\frac{276}{1084}=0.255$

(c) In this part we need probability that a person chosen at random is a female or someone who said yes .

   We define two events :

   Event A = person chosen is female

   Event B = person chosen said yes to the question

Probability of Event A happening = P(A) $=\frac{544}{1084}=0.502$

   Probability of Event B happening = P(B) $=\frac{516}{1084}=0.502=0.476$

  

Probability of a person chosen being female or someone who said yes = Probability of Event A or Event B occuring

   = P(AUB)

   (this is known as P(A or B)    , U is symbol known as union and used for or in statistics )

For calculating P(AUB) we have formula which is:

   P(AUB) = P(A) + P(B) - P(A$\cap$B) ( $\cap$ is the symbol known as intersection and means and in statistics)

   P(A$\cap$B) = Probability of Event A and Event B both occuring ,    which we calculated in second part

P(A$\cap$B) = 0.255

Required Probability = P(AUB) = 0.502 + 0.476 - 0.255 = 0.723

(d) This part is similar to (c) part as we need probability of Union.

We need probability that the chosen person is male or female .

   We define two events :

   Event A = person chosen being male

   Event B = person chosen being female

   P(A) = $\frac{540}{1084}=0.498$

P(B) = $\frac{544}{1084}=0.502$

   P( person being Female and male ) = P(A$\cap$B) = it is not possible anyone being both male and female ,so = 0

   P(A$\cap$B) = 0

  P(AUB) = P(A) + P(B) - P(A$\cap$B)

P(AUB) = P(Male or Female) = 0.498 + 0.502 = 1

   above result is obvious since any chosen has to be either male or female so probability of it occuring must be 1

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