Question

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?

Answer 1

Answer:

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  

Answer 2

Answer: 0.0304

Step-by-step explanation:

A z-test is a statistical test that is used to determine if two population means are different when the variances are known and the sample size is large. The test statistic is assumed to have a normal distribution, and nuisance parameters such as standard deviation should be known in order for an accurate z-test to be performed.

A z-statistic, also known as z-score, is a number representing how many standard deviations above or below the mean population a score derived from a z-test is.

Therefore,

sample size,n = 36

mean,u = 16 ounce

standard deviation,s = 0.2 pounds = 3.2 ounce

x = 15 ounce

z = (x-u) / (s/sqrt(n)) = -1.875

In conclusion, the probability that the amount dispensed by box will be increased is:

P(X<15) = P(Z<-1.875)

= 1 - P(Z<1.875)

= 1 - 0.9696

0.0304

The Answer is 0.0304.