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Lloyd’s Cereal company packages cereal in 1 pound boxes (16 ounces). A sample of 40 boxes is selected at random from the production line every hour, and if the average weight is less than 15 ounces, the machine is adjusted to increase the amount of cereal dispensed. If the mean for 1 hour is 1 pound and the σ = 0.1 pound, what is the probability that the amount dispensed per box will have to be increased?

Respuesta :

Answer:

Probability of amount dispensed per box is 0.0001

Step-by-step explanation:

Given data:

sample size, n = 40

mean[tex]\mu = 16 ounce[/tex]

standard deviation, s = 0.1 pounds

we know one pound equal to 16 ounce

therefore 0.1 pounds = 1.6 ounce

x = 15 ounce

[tex]z = \frac{(x-u)}{\frac{s}{\sqrt(n)}}[/tex]

z = -3.95

P(X<15) = P(Z<-3.95)

= 1 - P(Z< 3.952)    

= 1 - 0.9999

= 0.0001

so, Probability of amount dispensed per box is 0.0001

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