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the load of potato chips in a medium-dimension bag is mentioned to be 10 oz.. The volume that the packaging laptop places in these luggage is believed to have a normal mannequin with a median of ounces and a common deviation of oz. (circular to four decimal areas as essential.)
a) What fraction of all baggage sold are underweight?
b) probably the most chips are offered in "cut price packs" of 33 baggage. what is the chance that none of the 33 is underweight?
c) what's the likelihood that the imply weight of the 33 baggage is below the pointed out volume?
d) what's the chance that the mean weight of a 20-bag case of potato chips is below 10 oz?general Distribution:
A random variable X is declared to be following commonplace 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 \area N(\mu,\frac\sigma^2n). /eq
ordinary distribution is given by gauss and firstly it's used for modeling the error .
It is believed that the entire natural phenomenon follows regular distribution.answer and explanation:
it is for the reason that medium size chips is of 10 oz
The quantity that packaging computer put follow average distribution with mean and standard deviation
eq\mu= \\ \sigma= /eq
a) what fraction of bags are underweight less than 10 ounces
So we need to locate zscore after which the use of NORMSDIST (z) feature of MS Excel we will get the chance
eqP(X <10)=P(Z<\ /eq
b) P(None is underweight)=?
None is underweight skill all 33 aren't below 10 oz.
Let y denotes variety of underweight
P(None is underweight)=P(Y=0)=?
c) right here n=33
So we know that eq\bar x \sim N(\mu ,\frac\sigma^2n) /eq
for this reason eq\bar x /eq follow typical distribution with imply and commonplace deviation
eqP(\bar x <10)=P(Z<\ /eq
d) in a similar way here n=20
thus eq\bar x /eq follow usual distribution with mean and general deviation
eqP(\bar x <10)=P(Z<\ /eq