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a juice drink manufacture fills boxes advertised as containing 1l. a ma…

Question

a juice drink manufacture fills boxes advertised as containing 1l. a machine pours 1001 ml in each box with a standard deviation of 0.4 ml. any boxes containing less than 1000.2 ml are rejected. how many boxes will be rejected in a production run of 40,000 boxes?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 1000\), \(\mu=1001\), and \(\sigma = 0.4\).

$$z=\frac{1000 - 1001}{0.4}=\frac{-1}{0.4}=- 2.5$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X\lt1000)\). Looking up the z - score of \(-2.5\) in the standard normal distribution table, \(P(Z\lt - 2.5)=0.0062\)

Step3: Calculate the number of rejected boxes

If the total number of boxes \(n = 40000\), then the number of rejected boxes \(N=n\times P(X\lt1000)\)

$$N=40000\times0.0062 = 248$$

Answer:

\(248\)