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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: