Mathematics, 03.03.2020 23:33 fattypickeltoefungus

A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 P. M. on weekdays there is an average of four customers waiting in line. What is the probability that you visit Safeway today during this period and find?
a. No customers are waiting?
b. Four customers are waiting?
c. Four or fewer are waiting?
d. Four or more are waiting?

Answers: 3

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Answer from: reneecoutinho10

(a) The probability of no customers are waiting in a line is 0.01832.

(b) The probability of 4 customers are waiting in a line is 0.19537.

(d) The probability of 4 or more customers are waiting in a line during the visit is 0.56652.

Step-by-step explanation:

The number of customers waiting in a line between 4 PM and 7 PM (X) follows a Poisson distribution with parameter λ = 4.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-4}(4)^{x}}{x!} ;\ x=0, 1, 2,...$

(a)

Compute the probability that no customers are waiting in a line during the visit as follows:

$P(X=0)=\frac{e^{-4}(4)^{0}}{0!}=\frac{0.01832\times1}{1}=0.01832$

Thus, the probability of no customers are waiting in a line is 0.01832.

(b)

Compute the probability that 4 customers are waiting in a line during the visit as follows:

$P(X=4)=\frac{e^{-4}(4)^{4}}{4!}=\frac{0.01832\times256}{24}=0.19537$

Thus, the probability of 4 customers are waiting in a line is 0.19537.

(c)

Compute the probability that 4 or fewer customers are waiting in a line during the visit as follows:

P (X ≤ 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{-4}(4)^{0}}{0!}+\frac{e^{-4}(4)^{1}}{1!}+\frac{e^{-4}(4)^{2}}{2!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{4}}{4!}\\=0.01832+0.07326+0.14653+0.19537+0.19537\\=0.62885$

Thus, the probability of 4 or fewer customers are waiting in a line is 0.62885.

(d)

Compute the probability of 4 or more customers are waiting in a line during the visit as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-4}(4)^{0}}{0!}+\frac{e^{-4}(4)^{1}}{1!}+\frac{e^{-4}(4)^{2}}{2!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{3}}{3!}\\=1-0.01832-0.07326-0.14653-0.19537\\=0.56652$

Thus, the probability of 4 or more customers are waiting in a line during the visit is 0.56652.

Answer from: Quest

I’m not absolutely sure but i think it’s 7x6x5x4x3x2=5040 because as the first character you can take any of the 7 given then you are left with 5 then 6 then 4 etc.

Answer from: Quest

Do the math. she originally had 50 dollars on her gift card and she had 12 dollars left on her gift card after she finished her shopping spree. if you subtract 50 minus 12 you get thirty eight dollars as your answer. so therefore, ciara spent 38 dollars total o

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A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that betw...

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