Mathematics, 08.03.2021 23:20 emmahirstein

Assume that each time a metal detector at an airport signals, there is a 25% chance that the cause is change in the passenger's pocket. During a given hour, 15 passengers are stopped because of a signal from the metal detector. a) Find the probability that at least 3 passengers will have been stopped due to change in their pockets. b) If 15 passengers are stopped by the detector, would it be unusual for none of these to have been stopped due to change in the pocket

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Answer from: AbbyNeil

a) 0.7639 = 76.39% probability that at least 3 passengers will have been stopped due to change in their pockets.

b) 0 is less than 2.5 standard deviations from the mean, so it would not be unusual.

Step-by-step explanation:

For each passenger, there are only two possible outcomes. Either they will have been stopped due to a change in the pocket, or not. Passengers are independent. This means that 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.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A measure is considered unusual if it is more than 2.5 standard deviations of the mean.

25% chance that the cause is change in the passenger's pocket.

This means that p = 0.25

During a given hour, 15 passengers are stopped because of a signal from the metal detector.

This means that n = 15

a) Find the probability that at least 3 passengers will have been stopped due to change in their pockets.

This is

$P(X \geq 3) = 1 - P(X < 3)$

In which

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

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.2361 = 0.7639$

0.7639 = 76.39% probability that at least 3 passengers will have been stopped due to change in their pockets.

b) If 15 passengers are stopped by the detector, would it be unusual for none of these to have been stopped due to change in the pocket

First we find the mean and the standard deviation:

E(X) = np = 15*0.25 = 3.75

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15*0.25*0.75} = 1.68$

3.75 - 2.5*1.68 = -0.45

0 is less than 2.5 standard deviations from the mean, so it would not be unusual.

Answer from: Quest

rick has $1400 and sally has $1500

step-by-step explanation:

to find amount of money rick has after 10 years :

principal = $1000 , time = 10 years , rate = 4% annually

$\text{simple interest} = \frac{principal\times rate\times time}{100} \\\\\text{simple interest} = \frac{1000\times 4\times 10}{100}=\$400\\\\amount = principal + interest\\amount = 1000 + 400 = \$1400$

to find amount of money sally has after 10 years :

her friend gives to pay her $50 per year, therefore money she has after 10 years = 50 × 10 = $500

also, sally gets $1000 after 10 years,

therefore, total amount of money sally has after 10 years = 500 + 1000 = $1500

so, after 10 years : rick has $1400 and sally has $1500

Answer from: Quest

step-by-step explanation:

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