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The video states that 78% of people carry less than $50 in cash; 40% carry less than $20 in cash, and 9% carry no cash. Suppose you have a random sample of 100 people and you check how much cash each person is carrying. Assuming that the percentages stated in the video are correct for all adult Americans, discuss how it could occur that 12 people in your sample are carrying no cash while only 75 people in your sample are carrying less than $50 in cash.

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mirianmoses

Answer:

The probability that 12 people in your sample are carrying no cash is 0.0712

Step-by-step explanation:

n = 100

p(no cash) = 0.09

x = 12

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 12) = 0.074.

The probability that 12 people in your sample are carrying no cash is 0.074.

n = 100

p(less than 50) = 0.78

x = 75

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 75) = 0.0712

The probability that 12 people in your sample are carrying no cash is 0.0712

fichoh

Using the principle of binomial probability, the required probabilities to the problems posed in the question are ; 0.1299 and 0.071 respectively.

Recall :

  • P(x = x) = nCx * p^x * q^(n-x)
  • n = number of trials = 12

1.)

Probability that 12 people are carrying no cash :

[tex] 100C12 \times 0.09^{12} \times (1 - 0.09)^{100-12} [/tex]

[tex] 100C12 \times 0.09^{12} \times 0.91^{82} [/tex]

[tex] = 0.1299 [/tex]

2.)

Probability that 75 people are carrying less than 50 :

[tex] 100C75 \times 0.78^{75} \times (1 - 0.78)^{100-75} [/tex]

[tex] 100C75 \times 0.78^{75} \times 0.22^{25} [/tex]

[tex] = 0.071 [/tex]

Therefore, the probability that 85 out of the 100 samples are carrying less than $50 in cash is 0.071.

Learn more : https://brainly.com/question/12474772

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