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Lloyd's Cereal company packages cereal in 1 pound boxes (16 ounces). A sample of 36 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 standard deviation is 0.1 pound, what is the probability that the amount dispensed per box will have to be increased?

Respuesta :

ajeigbeibraheem

Answer:

0.0001

Step-by-step explanation:

Given that:

the sample (n) = 36  boxes

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

standard deviation [tex]\sigma[/tex] = 0.1 pounds

= 1.6 ounce

x = 5 ounces

∴

[tex]Z = \frac{(x- \ \mu)}{\frac{ \sigma }{ \sqrt{n} } }[/tex]

[tex]Z = \frac{(15- \ 16)}{\frac{ 1.6}{ \sqrt{36} } }[/tex]

[tex]Z= -3.75[/tex]

P ( X< 15) = P (Z < - 3.75)

= 1 - P (Z < 3.75)

= 1 - 0.9999

= 0.0001  

The  probability that the amount dispensed per box will have to be increased is 0.0001

Calculation of the probability:

Since

the sample (n) = 36  boxes

mean  = 16 ounce

standard deviation  = 0.1 pounds

= 1.6 ounce

And,

x = 5 ounces

Now for determining the probability first we have to find the z value

So,

Z value is

[tex]= (15-16) \div 1.6\div \sqrt36[/tex]

= -3.75

Now

The probability is

P ( X< 15) = P (Z < - 3.75)

= 1 - P (Z < 3.75)

= 1 - 0.9999

= 0.0001  

Hence, The  probability that the amount dispensed per box will have to be increased is 0.0001

Learn more about probability here: https://brainly.com/question/24613748

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