Question

In: Math

A recent survey was conducted to determine how people consume their news. According to this​ survey,...

A recent survey was conducted to determine how people consume their news. According to this​ survey, 60​% of men preferred getting their news from television. The survey also indicates that 42​% of the sample consisted of males. ​ Also, 42​% of females prefer getting their news from television. Use this information to answer the following questions.

a. Consider that you have 30000 ​people, given the information​ above, how many of them are​ male?
b. Again considering that you have 30000 ​people, how many of them are​ female?
c. Out of the males in your​ sample, how many of them prefer getting their news on​ television?
d. Out of the males in your​ sample, how many of them prefer getting their news​ on-line?
e. Out of the females in your​ sample, how many of them prefer getting their news on​ television?
f. Out of the females in your​ sample, how many of them prefer getting their news​ on-line?
g. How many people in your sample prefer getting their news from​ television?
h. How many people in your sample prefer getting their news from​ on-line?
nothing
i. Given that a person prefers getting their news on television​, what is the probability that the person is male ​(round to 3 decimal​ places)?

PLEASE SHOW ALL WORK

Solutions

Expert Solution

Assuming that the population prefer getting their news either online or on television

(a)

As 42% of the sample consists of males

Out of 30000 people, the number of male = 0.42*30000 = 12600

(b)

The number of females = (1 - 0.42)*30000 = 17400

(c)

Number of males in the sample who prefer getting their news on television = 0.60*12600 = 7560

(d)

Number of males in sample who prefer getting their news online

= (1 - 0.6)*12600 = 5040

(e)

Number of females who prefer getting their news on television

= 0.42*17400 = 7308

(f)

Number of females who prefer getting their news on line

= (1 - 0.42)*17400 = 10092

(g)

No of people in the sample who prefer getting their news from television = 7560 + 7308 = 14868

(h)

No of people in the sample who prefer getting their news from online

= 5040 + 10092 = 15132

(i)

Let A denote the event that the person is a male

and B denote the event that the person prefer getting news on television

Thus, P(A) = 0.42

P(B|A) = 0.60

P(B|A') = 0.42

Thus, P(B) = P(B|A)*P(A) + P(B|A')*P(A')

= 0.60*0.42 + 0.42*0.58 = 0.4956

The required probability = P(A|B)

= P(B|A)*P(A)/P(B) = 0.508


Related Solutions

At a recent concert, a survey was conducted that asked a random sample of 20 people...
At a recent concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the beginning of the year. The following data were collected: Age 62 57 40 49 67 54 43 65 54 41 # of Concerts 6 5 4 3 5 5 2 6 3 1 Age 44 48 55 60 59 63 69 40 38 52 # of Concerts 3 2 4 5 4 5...
A survey is conducted to find out whether people in metropolitan areas obtain their news from...
A survey is conducted to find out whether people in metropolitan areas obtain their news from television (Event T), a newspaper (Event N), or radio (Event R). The results show that 60% of people get news on television, 60% from newspapers, 50% from radio, 30% from television and newspapers, 25% from television and radio, 30% from newspapers and ratio, and 15% from television, newspapers and radio. a) What is the proportion of those who obtain news from television, but not...
5. Scenario: A survey is conducted to determine the average number of people living in a...
5. Scenario: A survey is conducted to determine the average number of people living in a household in Scranton. An interviewer asks a sample of 16 Scranton shoppers, selected randomly in the Viewmont Mall (no two living in the same household), how many people live in their household. The results are as follows: 1        1      1      2     2     2      2     2   2     3     3     3     4     4    5     6 Assume the distribution of the number in all households is...
According to a recent survey conducted at a local college, we found students spend an average...
According to a recent survey conducted at a local college, we found students spend an average of 19.5 hours on their smartphone per week, with a standard deviation of 3.5 hours. Assuming the data follows the normal distribution. a) How many percent of students in this college spend more than 15 hours on their smartphone per week? b) If we randomly select 12 students, what is the probability that the average of these students spending on the internet is more...
According to the recent survey conducted by Sunny Travel Agency, a married couple spends an average...
According to the recent survey conducted by Sunny Travel Agency, a married couple spends an average of $500 per day while on vacation. Suppose a sample of 64 couples vacationing at the Resort XYZ resulted in a sample mean of $490 per day and a sample standard deviation of $100. Develop a 95% confidence interval estimate of the mean amount spent per day by a couple of couples vacationing at the Resort XYZ.
In a recent survey, 2033 people were surveyed as to how many people preferred Tinder and...
In a recent survey, 2033 people were surveyed as to how many people preferred Tinder and how many people preferred Happn (a local dating website) and how many preferred Tinder in New York and Arizona. It was found that 1690/2033 people in New York liked using Tinder while 343/2033 preferred Happn and in Arizona, 1853/2033 people liked using Tinder while 180/2033 liked using Happn. Construct and interpret a 95% confidence interval for the difference between the two dating website preferences....
A survey was conducted that asked 1008 people how many books they had read in the...
A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.
A survey was conducted that asked 1008 people how many books they had read in the...
A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.
A survey was conducted that asked 1023 people how many books they had read in the...
A survey was conducted that asked 1023 people how many books they had read in the past year. Results indicated that x over bar x equals=15.4 books and s equals=18.2 books. Construct a 95​% confidence interval for the mean number of books people read. Interpret the interval. confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round...
A survey was conducted that asked 1019 people how many books they had read in the...
A survey was conducted that asked 1019 people how many books they had read in the past year. Results indicated that x overbarequals10.9 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval. Construct a 99​% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round to...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT