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ome Syllabus Content AssessmentsCommunicationResources LibraryDesignTutoring n 11 Quiz The Binomial Distribution 2:00:00 Jacoby Wright: Attempt 3 stions saved Question 5(1.2 points) The probability that a randomly selected individual in a certain community has made an online purchase is 0.38. Suppose that a sample of 13 people from the community is selected, what is the probability that at most 3 of them has made an online purchase? Write only a number as your answer. Round to 2 decimal places (for example 0.24) Do not write as a percentage Your Answer: Answer ed Response saved Response o Item Save Question 6 (1.2 points) At a major International Airport, 0.35 of the flights arrive on time. A sample of 11 flights is studied. What is the probability that more than 3 of them arrived on time? Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage Your Answer Question 7 Saved

Explanation / Answer

Solution(5)
Given in the question
P = 0.38
n = 13
We need to calculate P(X<=3 | n=13) = P(X=0)+P(X=1)+P(X=2)+P(X=3)
= 13C0*(0.38)^0 * (0.62)^13 + 13C1*(0.38)^1 * (0.62)^12 +13C2 *(0.38)^2 * (0.62)^11 + 13C3 *(0.38)^3 *(0.62)^10
= 0.0020 + 0.01593 + 0.0586 + 0.13171 = 0.2082
So there is 0.21 or 21% probability that at most 3 of them has made an online purchase.
Soluion(6)

Given in the question
p = 0.35
n = 11
P(X>3 | n=11) = 1- P(X=0) + P(X=1) +P(X=2) +P(X=3
P(X>3 | n=11) = 1 - 11C0 * (0.35)^0 *(0.65)^11 + 11C1 *(0.35)^1 *(0.65)^10 +11C2 *(0.35)^2 *(0.65)^9 + 11C3 *(0.35)^3 *(0.65)^8 = 1 - 0.00875 - 0.05185 - 0.1395 - 0.2254 = 1- 0.4255 = 0.5744
So there is 0.57 or 57% probability that more them 3 of them arrived on time.

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