Monogram Masters advertises on its website that 92% of customer orders are received within five working days. They performed an audit from a random sample of 200 of the 3,000 orders that month and it shows 175 orders were received on time.

Part A: Can we use a normal approximation? Explain. (5 points)

Part B: If Monogram Masters customers really receive 92% of its orders within five working days, what is the probability that the proportion in the random sample of 200 orders is the same as the proportion found in the audit sample or less? (5 points)

a yes

0.0094

Step-by-step explanation:

Part a

p = 0.92, n = 200

np = 184 > 10

n(1 − p) = 16 > 10

Therefore, normal approximation can be used.

μ = p = 0.92

σ = √(p (1 − p) / n)

σ = √(0.92 × 0.08 / 200)

σ = 0.0192

x = 175/200 = 0.875

z = (x − μ) / σ

z = (0.875 − 0.92) / 0.0192

z = -2.34

P(z < -2.34) = 0.0096

Part b

It is 99% that the proportion is not 0.92

Therefore, there is enough evidence at the 0.01 level, that monogram master ships less tha 92% of its order on time

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all