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the weight of potato chips in a medium-size bag is stated to be 10 ounces. The quantity that the packaging computer puts in these bags is believed to have a traditional mannequin with a mean of oz and a standard deviation of oz.. (circular to 4 decimal areas as necessary.)
a) What fraction of all baggage sold are underweight?
b) one of the most chips are bought in "bargain packs" of 33 bags. what's the likelihood that not one of the 33 is underweight?
c) what is the chance that the suggest weight of the 33 baggage is below the stated volume?
d) what's the chance that the suggest weight of a 20-bag case of potato chips is beneath 10 oz?usual Distribution:
A random variable X is declared to be following normal distribution with meaneq\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 \space \sim \area N(\mu,\frac\sigma^2n). /eq
usual distribution is given by using gauss and initially it's used for modeling the error .
It is assumed that the entire herbal phenomenon follows common distribution.reply and clarification:
it's on account that medium size chips is of 10 oz.
The amount that packaging computer put observe standard distribution with imply and ordinary deviation
eq\mu= \\ \sigma= /eq
a) what fraction of baggage are underweight lower than 10 oz.
So we should discover zscore after which the usage of NORMSDIST (z) feature of MS Excel we can get the likelihood
eqP(X <10)=P(Z<\ /eq
b) P(None is underweight)=?
None is underweight potential all 33 don't seem to be below 10 oz
Let y denotes variety of underweight
P(None is underweight)=P(Y=0)=?
c) right here n=33
So we recognize that eq\bar x \sim N(\mu ,\frac\sigma^2n) /eq
consequently eq\bar x /eq observe regular distribution with mean and typical deviation
eqP(\bar x <10)=P(Z<\ /eq
d) in a similar way right here n=20
therefore eq\bar x /eq observe typical distribution with imply and ordinary deviation
eqP(\bar x <10)=P(Z<\ /eq