Mathematics: 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?

Lloyd s Cereal company packages cereal in 1 pound

ciel8809

22.01.2022

0.0001

Step-by-step explanation:

Given that:

the sample (n) = 36 boxes

mean $( \mu)$ = 16 ounce

standard deviation $\sigma$ = 0.1 pounds

= 1.6 ounce

x = 5 ounces

∴

$Z = \frac{(x- \ \mu)}{\frac{ \sigma }{ \sqrt{n} } }$

$Z = \frac{(15- \ 16)}{\frac{ 1.6}{ \sqrt{36} } }$

Z= -3.75

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

= 1 - P (Z < 3.75)

= 1 - 0.9999

= 0.0001

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AADJ4

19.10.2022

Third option is correct.

Step-by-step explanation:

Since we have given that

Initial value of the property = P

Rate of growth = 275%

So, we need to find the property's current value:

We will use "compound interest ":

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