Question

A local courier service reports that 82% of bulk parcels within the same city are delivered within two days. Six parcels are randomly sent to different locations. a. What is the probability that all six arrive within two days? (Round the final answer to 4 decimal places.) Probability b. What is the probability that exactly five arrive within two days? (Round the final answer to 4 decimal places.) Probability c. Find the mean number of parcels that will arrive within two days. (Round the final answer to 2 decimal place.) Number of parcels d-1. Compute the variance of the number that will arrive within two days. (Round the final answer to 3 decimal places.) Variance d-2. Compute the standard deviation of the number that will arrive within two days. (Round the final answer to 4 decimal places.) Standard Deviation

Answer 1

a)

If using a calculator, you can enter trials = 6, p=0.82, and X = 6 into a binomial probability distribution function (PDF). If doing this by hand, apply the binomial probability formula: P(X) = ().pt. (1–p)*-* The binomial coefficient, is defined by () n! X!(n - X)! The full binomial probability formula with the binomial coefficient is P(X) = n! .p\.(1 - p)"-X X!(n - X)! where n is the number of trials, p is the probability of success on a single trial, and X is the number of successes. Substituting in values for this problem, n=6, p=0.82, and X = 6. 66 P(6) 6! -0.826 - (1 - 0.82) 6!(6-6)! Evaluating the expression, we have P(6)=0.3040

b)

P(5) Probability of exactly 5 successes If using a calculator, you can enter trials = 6, p=0.82, and X = 5 into a binomial probability distribution function (PDF). If doing this by hand, apply the binomial probability formula: P(X) n X p\.(1 - p)n-X The binomial coefficient, () is defined by n n! X!(n - X)! X The full binomial probability formula with the binomial coefficient is P(X) n! .p*. (1 - p)-X X!(n - X)! where n is the number of trials, p is the probability of success on a single trial, and X is the number of successes. Substituting in values for this problem, n=6, p=0.82, and X = 5. P(5) 6! -0.825 - (1 -0.82)65 5!(6-5)! Evaluating the expression, we have P(5) = 0.4003

c)

Mean u = n.p = (6) 41 50 123 25 4.92.

d1)

Variance o2 = n.p.(1 - p) = (6). (*): (1-() 1107 1250 0.8856.

d2)

31246 Standard deviation o = vm.p: (1–p) = 6) (1-0) 50 0.94101