Answer to Question #252180 in Statistics and Probability for Ziyee

Question

Accuracy in taking orders at a drive-through window is important for fast-food chains.

Each month, QSR Magazine, www.qsrmagazine.com, publishes the results of its

surveys. Accuracy is measured as the percentage of orders that are filled correctly. In

a recent month, the percentage of orders filled correctly at McDonald’s was

approximately 91%. Suppose that you and ten friends go to the drive through window

at McDonald’s and independently place orders,

i. what is the probability that at least 7 orders will be filled correctly?

ii. what is the probability that between 3 and 9 orders will be filled

correctly?

iii. what is the probability that not more than 5 orders will NOT be filled

correctly?

Answer 1

This situation can be desribed with binomial distribution X = Bin(10, 0.91), where X - number of succescfully filled orders

(i) The probability that at least 7 orders will be filled correctly found as P(X ≥ 7)

$P(X≥7) = P(X=7) +P(X=8)+P(X=9)+P(X=10) = 0.0452+0.1714+$

$+0.3851 + 0.3894 = 0.9911$

(ii) The probability that between 3 and 9 orders will be filled correctly is P(3 ≤ X ≤ 9)

$P(3≤X≤9) = P(X=3)+...+P(X=9) = 0.6105\\$

(iii) The probability that not more than 5 orders will NOT be filled correctly is P(X≥5)

$P(X≥5) = P(X=5) + ... P(X=10)= 0.9998$