Monogram masters advertises on its website that 92% of customer orders are received within five working days. they perform 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 distribution? explain.
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?

all below

Step-by-step explanation:

Part A: The hypothesized value of proportion is P=0.92. So for sample size n=200, the standard deviation of distribution is σ=sqrt/px(1-p) divided by n = 0.0192

For given observation, the sample proportion p= 175/200 = 0.875

So the test statistic is Z= p-P/σ = 0.875 - .92/ 0.0192 = -2.34

Required probability = P(Z< -2.34)=0.0096

Therefore, the probability that in the simple random sample of 200 orders, is as small as the proportion in the auditors' sample or smaller is just less than 1%.\

Part B: It is 99% probable that the proportion is not 0.92. There is enough evidence, at the 0.01 level, that Monogram Masters ships less than 92% of its orders on time.

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4

answer: 5

step-by-step explanation: