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2 years ago

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In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg

ebra and 3 study neither algebra nor statistics. Each student studies at least one of the two subjects. 1. Illustrate the information on a venn diagram. 2. How many students study a. Algebra b. Statistics

1 answer:

Vladimir [108]2 years ago

6 0

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

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Step-by-step explanation:

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bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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