Question

Need statistics help again.

When a truckload of apples arrives at a packing plant, a random sample of 200 is selected and examines for bruises, discolorations etc. The whole truckload will be rejected if more than 6% of the sample is unsatisfactory. Suppose that in fact 9% of the apples on the truck do not meet the desired standard. What is the probability that the shipment will be accepted anyway?

Answer 1

This can be solved by using the normal approximation to the binomial distibution.
The mean is np=200 * 0.09 = 18
The standard deviation is
\[\sqrt{np(1-p)}=\sqrt{200\times 0.09\times 0.91}=4.04722\]
Now we need to find the z-score for 12 which corresponds to 6% of the apples being unsatisfactory.
\[z=\frac{X-\mu}{\sigma}=\frac{12-18}{4.04722}\]
The cumulative probability for the z-score is found by using a standard normal distribution table.

Answer 2

Thank you very much!

Answer 3

You're welcome :)