The Incredible Process of Packaging Potato Chips ...vertical form fill seal packaging machines grain packaging equipment sealing
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the load of potato chips in a medium-measurement bag is cited to be 10 oz. The volume that the packaging computer places in these luggage is believed to have a normal model with an average of ounces and a typical deviation of ounces. (circular to four decimal places as essential.)
a) What fraction of all bags sold are underweight?
b) some of the chips are offered in "cut price packs" of 33 baggage. what's the chance that none of the 33 is underweight?
c) what is the probability that the suggest weight of the 33 bags is under the cited volume?
d) what's the likelihood that the imply weight of a 20-bag case of potato chips is beneath 10 oz.?common Distribution:
A random variable X is declared to be following general distribution with implyeq\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
standard distribution is given by gauss and originally it's used for modeling the error .
It is believed that the entire natural phenomenon follows common distribution.answer and explanation:
it is for the reason that medium dimension chips is of 10 ounces
The quantity that packaging computer put follow regular distribution with mean and general deviation
eq\mu= \\ \sigma= /eq
a) what fraction of bags are underweight below 10 oz.
So we deserve to discover zscore after which the use of NORMSDIST (z) characteristic of MS Excel we are able to 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) right here n=33
So we comprehend that eq\bar x \sim N(\mu ,\frac\sigma^2n) /eq
consequently eq\bar x /eq observe normal distribution with mean and average deviation
eqP(\bar x <10)=P(Z<\ /eq
d) similarly right here n=20
as a result eq\bar x /eq comply with normal distribution with imply and typical deviation
eqP(\bar x <10)=P(Z<\ /eq