Probability of Acceptance for Container Loads at Shedz

[Paulo2014]

Homework Statement

(a) Shedz buys its trays in pallets by the container load. As each container arrives, n trays are examined
to assess whether or not the load of trays is of acceptable quality. The load is accepted if there is no
more than one defective tray in the sample.
(i) For two successive loads, 1% and 2% of the trays were defective, respectively.
If n = 50 for each load, find the probability that at least one of these two loads was accepted.
(ii) If 1% of the trays are defective in a load, find the largest value of n which ensures there is
approximately a 94% chance of the load being accepted.
(b) Each pallet consists of 180 trays stacked in 30 levels, with six trays arranged as a 3 by 2 rectangle at
each level.
(i) Describe how cluster sampling could be used to select 24 trays from a container load of 20
pallets.
(ii) Give one advantage and one disadvantage of using cluster sampling in this situation,
compared with simple random sampling.

Homework Equations

The Attempt at a Solution

I'm completely stuck...

 

[Nino MMP]

I think the first part has something to do with binomial probability, but I'm not sure how to apply it.

 

[Architeuthis Dux]

I can provide some insights and suggestions for the given content.

(a) The probability of acceptance for container loads at Shedz depends on the number of defective trays in the sample. In order to calculate the probability, we need to use the binomial distribution formula.

(i) For the first load, n = 50 and the probability of a defective tray is 1%. Therefore, the probability of accepting the load is:

P(accept) = 1 - P(defective) = 1 - (0.01) = 0.99

For the second load, n = 50 and the probability of a defective tray is 2%. Therefore, the probability of accepting the load is:

P(accept) = 1 - P(defective) = 1 - (0.02) = 0.98

Now, to find the probability that at least one of these two loads was accepted, we can use the formula for the union of two events:

P(at least one load accepted) = P(accept first load) + P(accept second load) - P(both loads accepted)

= 0.99 + 0.98 - (0.99)(0.98)

= 0.9998

Therefore, there is a 99.98% chance that at least one of these two loads was accepted.

(ii) To find the largest value of n which ensures a 94% chance of acceptance, we can use the inverse binomial distribution function. We know that the probability of acceptance is 0.94 and the probability of a defective tray is 0.01. Therefore, we can solve for n using the following equation:

0.94 = 1 - (0.01)^n

n = ln(0.06)/ln(0.99)

n ≈ 241 trays

Therefore, if n is at least 241 trays, there is approximately a 94% chance of the load being accepted.

(b) Cluster sampling is a sampling technique in which the population is divided into clusters and a random sample is taken from each cluster.

(i) In this situation, we can divide the container load into 20 clusters, each cluster representing one pallet. Then, we can randomly select 24 trays from each cluster to get a total sample size of 480 trays.

(ii) The advantage of using cluster sampling is that it is more