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cricket20 [7]
4 years ago
15

_________________represents the amount or quantity of something. It could be discrete or continuous. Examples are number of fema

le students in CMU, heights of students in a class
Mathematics
1 answer:
Evgen [1.6K]4 years ago
4 0

Answer:

variable

Step-by-step explanation:

A variable is a symbol that represents a single quantity but it is called a variable because the quantity/value/amount is unknown and it can take on (can be) any possible value (number). The word "variable" itself means "values that vary". This means that the value is unknown as it stands but it can take on any value or outcome.

You can call this variable anything, in Maths we often use letters.

eg. X is the variable we're interested in. X can take on any number. Let X be the number of people living in a house. For this example lets say X = 8. ie. there are 8 people living in the house.

- in this case x is discrete.

There are two types of variables: discrete and continuous.

<u>Discrete variable: </u>

- countable; nothing in between, can only take on a finite number of values

This is a variable that ca take on any countable number along an interval. A way of thinking about a discrete variable is to think of is as a count. Anything that you can count as whole numbers.

Eg. X is a discrete variable whose <u>value lies along the interval [2;3]</u>. This means the value of  X can ONLY be 2 or 3.

Examples include: number of apples, number people, toes, trees, real numbers on an interval (2,3,4,5,6..), ages of a group of friends, rounded of values etc.

<u>Continuous: </u>

- can take on an infinite number of values; always something in between; impossible to count realistically

A continuous variable means that it can take on any value along an interval between to whole numbers. It is continuous along that interval.

Eg. X is a continuous variable whose <u>value lies along the interval [2;3]</u>. This means that it can take on absolutely any value that lies between 2 and 3: 2.00000000001 or 2.9999999 or even just 2.5. The options are endless.

Examples include: height, weight, temperature etc.

<u>SIDE NOTE: </u>

This can be extended to a random variable which follows the same concept except applies it to probability. You shouldn't be confused with this because while a variable represents a single quantity, a random variable represents the outcome of a process: a value that follows a specific probability distribution which is determined from a probability function. Therefore, it can take on multiple values. Look into this only if necessary.

:)

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Answer:

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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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Answer:

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