Students at a certain school can enroll in one elective course: painting, theater, chior, or band. This

9 answers
Question:

Students at a certain school can enroll in one elective course: painting, theater, chior, or band. This two-way frequency table gives the number of male and female studetns enrolled in eah class. Determine the conditional realtive that a student is male, given that the student is enrolled in chior.

Answers

45.7

Step-by-step explanation:

I took the test and this is what the right answer is

(A) it is impossible for the mean score to be 8 because the mean is the average and there is no 8 even on the table or on an ordinary die numbered 1-6, therefore You cannot get an average of 8 if there is no existence of an 8 on the die or in the score.

(B) To get the mean of the score, you should list the numbers and put how many numbers the frequency is. (Example: 1,1,1,1,1,1,1.) because on the chart it says the 1 has a frequency of 7. So your list should look like this:
1,1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,5,5,6,6,6,6
So then you add them all up:
The 1’s=7
The 2’s=10
The 3’s=12
The 4’s=16
The 5’s=30
The 6’s=24
Then you get a grand total of 99, then you divide for how many numbers you had in your list which is 30, then you divide 99 by 30 (99/30=x) and when you do the math x should equal 3.3, so the mean of the scores is 3.3.

The probability of having high cholesterol = 0.496

Step-by-step explanation:

Given frequency data table

                                                         cholesterol level

                                          Normal           High              Total

Heart diseases      Yes       11                   223                  234

                               No       367                149                   516

                             Total        378               372                 750

Given data

The Cholesterol level is high  n ( High cholesterol) = 372

Total n( S) = 750

The probability of having high cholesterol level

                           P( E) = [tex]\frac{n(E)}{n(S)} = \frac{372}{750} = 0.496[/tex]

[tex]Mean = 69[/tex]

[tex]Mode = 66[/tex]

[tex]Median = 69[/tex]

Step-by-step explanation:

Given

The frequency table

Required

Determine the mean, median and mode

Calculating Mean

[tex]Mean = \frac{\sum fx}{\sum f}[/tex]

Where

fx = product of frequency and inches

f = frequency

So;

[tex]Mean = \frac{63 * 2 + 65 * 1 + 66 * 4 + 67 * 3 + 68 * 1 + 69 * 2 + 70 * 2 + 71 * 1 + 72 * 3 + 74 * 2 + 75 * 2}{2 + 1 + 4 + 3 + 1 + 2 + 2 + 1 + 3 + 2 + 2}[/tex]

[tex]Mean = \frac{1587}{23}[/tex]

[tex]Mean = 69[/tex]

Calculating Mode

[tex]Mode = 66[/tex]

Because it highest frequency of 4

Calculating Median

[tex]Median = \frac{\sum f}{2}th\ position[/tex]

[tex]Median = \frac{2 + 1 + 4 + 3 + 1 + 2 + 2 + 1 + 3 + 2 + 2}{2}th\ position[/tex]

[tex]Median = \frac{23}{2}th\ position[/tex]

[tex]Median = 11.5th\ position[/tex]

Approximate

[tex]Median = 12th\ position[/tex]

At this point we, need to get the cumulative frequency (CF)

Inches ---- Frequency ---- CF

63 -------------2------------------2

65 -------------1------------------3

66 -------------4------------------7

67 -------------3------------------10

68 -------------1------------------11

69 -------------2------------------13

70 -------------2------------------15

71 -------------1------------------16

72 -------------3------------------19

74 -------------2------------------21

75 -------------2------------------23

From the above table

Since the median fall in the 12th position, then we consider the following data

69 -------------2------------------13

because it has a CF greater than 12

Hence;

[tex]Median = 69[/tex]

A table divided into cells by category with counts for each category in each cell. For example, let's say you were counting the number of cars and trucks that drove down a road each day over a 5-day week. Your categories would be vehicle and day. You could summarise this as a frequency table:

Step-by-step explanation:

The correct answer is 60%. Just took this quiz.

The likelihood that he or she has a pet given that a student does not have a sibling is:

Option: D ( 60%)

Step-by-step explanation:

Let A denote the event that a student does not have a sibling.

Let B denote the event that a students has a pet.

Then A∩B denotes the event that the student does not have a sibling but has a pet.

Let P denote the probability of an event.

Then, we are asked to find:

P(B|A)

We know that:

[tex]P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}[/tex]

Also from the table we have:

P(A)=0.25

and P(A∩B)=0.15

Hence,

[tex]P(B|A)=\dfrac{0.15}{0.25}\\\\P(B|A)=\dfrac{3}{5}\\\\P(B|A)=0.6[/tex]

which in percentage is:

[tex]0.6\times 100=60\%[/tex]

Hence, the likelihood is:

60%

The likelihood that he or she has a pet given that a student does not have a sibling is:

Option: D ( 60%)

Step-by-step explanation:

Let A denote the event that a student does not have a sibling.

Let B denote the event that a students has a pet.

Then A∩B denotes the event that the student does not have a sibling but has a pet.

Let P denote the probability of an event.

Then, we are asked to find:

P(B|A)

We know that:

[tex]P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}[/tex]

Also from the table we have:

P(A)=0.25

and P(A∩B)=0.15

Hence,

[tex]P(B|A)=\dfrac{0.15}{0.25}\\\\P(B|A)=\dfrac{3}{5}\\\\P(B|A)=0.6[/tex]

which in percentage is:

[tex]0.6\times 100=60\%[/tex]

Hence, the likelihood is:

60%

Frequency tableA frequency table is a table that summarizes a data set by stating the number of times each value occurs within the data set. HistogramA histogram is a display that indicates the frequency of specified ranges of continuous data values on a graph in the form of immediately adjacent bars

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