Question

Seams Personal advertises on its website that 95% of customer orders are received within four working days. They performed an audit from a random sample of 500 of the 6,000 orders received that month and it shows 470 orders were received on time. Part A: Can we use a normal approximation? Explain. (5 points) Part B: If Seams Personal customers really receive 95% of their orders within four working days, what is the probability that the proportion in the random sample of 500 orders is the same as the proportion found in the audit sample or less? (5 points) (10 points)

Answer 1

Answer:

a. Yes(n=500>=5, n(1-p)=25>=5)

b. 0.15241

Step-by-step explanation:

a. A normal approximation to the binomial can be used  $n\geq$5 and n(1-p)>=5:

#We calculate our p as follows:

$\hat p$=x/n=470/500=0.94

n=500

n(1-p)=500(1-0.95)=25

Hence, we can use the normal approximation.

b. This is a normal approximation.

-Given that p=0.95(95%)

-We verify if our distribution can be approximated to a normal:

$np=0.95\times 500=475\\n(1-p)=500(1-0.95)=25\\\\np\geq 5,\ n(1-p)\geq 5$

Hence, we can use the normal approximation of the form:

$P_{bin}(k,n,p)->N(\mu,\sigma^2)\left \{ {{\mu=np=475} \atop {\sigma=\sqrt{np(1-p)}=4.8734}} \right. \\\\\\P_{bin}(k\leq 470)\approx P_{norm}(x\leq 470.5)=P_{norm}(z\leq \frac{470-475}{4.8734})\\\\P_{norm}(z\leq -1.0260)=0.15241$

Hence, the probability of the sample proportion  is the same as the proportion of the sample found is 0.15241