Probability

A quality control audit has been devised to check on the sampling procedure when a truckload of potatoes arrive at a packing plant. A random sample of 250 is selected and examined for bruises and other defects. The whole truckload will be rejected if 5% of the sample is unsatisfactory. Determine the probability that the shipment will be accepted anyway if the load were to have 8% of the potatoes on the truck not meeting the desired standard.State assumptions and conditions and if they are met.

am i going to use below formula and using the hypothesis testing?

z= p(hat) - p / Sd(p-hat)

Please advise?

This is how I figured it out!

n=250
Sample mean =n*p=250*0.08=20
Population mean=n*p=250*0.05=12.5
Sample SD =Sqrt(n*P*q)=sqrt(250*0.05*0.95)=3.446
Since the sample size is large enough and np and npq are greater than 5 the assumption of
binomial approximation to normal is met.

You want to find the probability that the shipment will be accepted anyway if the load were 8% of the
potatoes on the truck not meeting the desired standard.
P(X<20)=P(X-Mean/SD<(20-12.5)/3.446)=P(Z<7.5/3.446)=P(Z<2.176)
=0.5+P(<Z<2.18)=0.5+0.48537=0.98537
So the probability that the shipment will be accepted anyway if the load were 8% of the
potatoes on the truck not meeting the desired standard is 0.9854.
Hope this helps!