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Ilia_Sergeevich [38]
3 years ago
13

Need Help ASAP!!!!!!

Mathematics
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

a) 62%

b) 44%

Step-by-step explanation:

Beforehand we have the information that the amount of shoppers, let`s call it n= 150.

To answer the first question, we need to know the amount of shoppers that do not possess an email account: this is the sum= 18`+75 (both the shoppers without and with access to a computer at home buth without an email). Lets call it a.

The percentage is defined by:

%of shoppers that dont have and email account= a/n*100= (18+75)*100/150= 62%

The next question is the percentage of shoppers with computer access at home, thus the sum of 48+18, both who have and dont have an email account but the have computer access at home. Lets call it b.

So

%of shoppers that have computer access at home = b/n*100= (48+18)*100/150= 44%

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The domain is all real numbers.

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Consider the following binomial experiment. The probability that a green jelly bean is chosen at random from a large package of
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There is a 5.79% probability that at most 2 jelly beans are chosen.

Step-by-step explanation:

For each jelly bean chosen, there are only two possible outcomes. Either it is green, or it is not. This is why we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

There are 13 jelly beans, so n = 13.

The probability that a green jelly bean is chosen at random from a large package of jelly beans is 2/5. This means that p = 0.4.

If Sally chooses 13 jelly beans, what is the probability that at most two will be green jelly beans?

This is

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

In which

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

P(X = 0) = C_{13,0}.(0.4)^{0}.(0.6)^{13} = 0.0013

P(X = 1) = C_{13,1}.(0.4)^{1}.(0.6)^{12} = 0.0113

P(X = 1) = C_{13,2}.(0.4)^{2}.(0.6)^{111} = 0.0453

So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0013 + 0.0113 + 0.0453 = 0.0579.

There is a 5.79% probability that at most 2 jelly beans are chosen.

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