Question

A tuna producer claims that its cans of tuna contain a standard amount of tuna. It has been determined that of 200 cans manufactured, 92% contain the standard amount. An inspector will only accept a shipment if at least 175 out of 200 sample cans contain more than the standard amount. What is the probability that the shipment is accepted?

Answer 1

Let $X=$ the number of cans that contain the standard amount of tuna: $X\sim Bin (n, p).$

Given $n=200, p=0.92.$

If the binomial probability histogram is not too skewed, $X$ has approximately a normal distribution with $\mu=np$ and $\sigma=\sqrt{npq}.$

In practice, the approximation is adequate provided that both $np\geq 10$ and $nq\geq 10,$ since there is then enough symmetry in the underlying binomial

distribution.

$np=200(0.92)=184\geq 10$

$nq=n(1-p)=200(1-0.92)=16\geq 10$

Hence we can use Normal approximation to the Binomial: $X\sim N(\mu, \sigma^2)$

$$\mu=np=200(0.92)=184$$

$$\sigma^2=npq=200(0.92)(1-0.92)=14.72$$

Use the continuity correction factor

$$P(X\geq 175)=P(X>175-0.5)$$

$$=1-P(X\leq174.5)=1-P(Z\leq\dfrac{174.5-184}{\sqrt{14.72}})$$

$$\approx1-P(Z\leq-2.4761)\approx0.9934$$

The probability that the shipment is accepted is 0.9934.