Multi-head Weight Machine For Sachets|Chips Snack Weighing ...
The Multi-head Weight Machine For Sachets|Chips Snack Weighing ... we manufacture has a good reputation both domestically and internationally.To be productive, we maintain an attitude of excellence.It is easy to provide chips packing machine products in reasonable price, because our factory has matured craft and professional mechanic.We look forward to working with customers around the world.The quality of our products will surely satisfy you, and our service will reassure you even more.I wish you every success in your career and a great deal of money.
the weight of potato chips in a medium-size bag is stated to be 10 oz.. The quantity that the packaging computing device places in these luggage is believed to have a standard mannequin with an average of oz and a standard deviation of ounces. (circular to four decimal locations as needed.)
a) What fraction of all baggage offered are underweight?
b) one of the crucial chips are offered in "bargain packs" of 33 luggage. what's the likelihood that not one of the 33 is underweight?
c) what's the probability that the suggest weight of the 33 bags is under the cited quantity?
d) what is the likelihood that the imply weight of a 20-bag case of potato chips is beneath 10 oz.?usual Distribution:
A random variable X is said to be following typical distribution with suggesteq\mu /eq and variance eq\sigma^2 /eq if its distribution is given as eqf(x|\mu,\sigma^2)=\frac1\sqrt2\pi \sigma^2e^-\frac(x-\mu)^22\sigma^2 \qquad , -\infty \leq X \leq \infty \\ \bar x \house \sim \space N(\mu,\frac\sigma^2n). /eq
commonplace distribution is given through gauss and at the start it's used for modeling the error .
It is thought that the entire herbal phenomenon follows regular distribution.reply and rationalization:
it's on account that medium size chips is of 10 oz
The quantity that packaging computer put observe ordinary distribution with imply and commonplace deviation
eq\mu= \\ \sigma= /eq
a) what fraction of bags are underweight less than 10 oz.
So we should locate zscore after which the usage of NORMSDIST (z) feature of MS Excel we will get the likelihood
eqP(X <10)=P(Z<\ /eq
b) P(None is underweight)=?
None is underweight skill all 33 don't seem to be beneath 10 oz.
Let y denotes variety of underweight
P(None is underweight)=P(Y=0)=?
c) here n=33
So we know that eq\bar x \sim N(\mu ,\frac\sigma^2n) /eq
as a consequence eq\bar x /eq follow average distribution with imply and ordinary deviation
eqP(\bar x <10)=P(Z<\ /eq
d) in a similar fashion here n=20
accordingly eq\bar x /eq comply with standard distribution with imply and normal deviation
eqP(\bar x <10)=P(Z<\ /eq