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Alina [70]
3 years ago
10

Laboratory Worksheet

Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0
Heyyyyyyyyyyy Hii what’s up
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A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol
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

8 0
3 years ago
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