The amount of cereal that can be poured into a small bowl varies with

Michael Dennis 2021-11-15 Answered
The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. a) How much more cereal do you expect to be in the large bowl? b) What 's the standard deviation of this difference? c) If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d) What are the mean and standard deviation of the total amount of cereal in the two bowls? e) If the total follows a Normal model, what's the probability you poured out more than 4.5 ounces of cereal in the two bowls together? f) The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and the standard deviation.

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Expert Answer

Oung1985
Answered 2021-11-16 Author has 1662 answers

If X is the random variable representing the small bowl and Y the random variable representing the large bowl, then \(\displaystyle{Y}-{X}\) represents the difference between the large and small bowl.
a) The expected value changes the same as the random variables:
\(\displaystyle\mu-\mu_{{Y}}-\mu_{{X}}={2.5}-{1.5}={1}\)
b) The variance changes by the square of the coefficients and the standard deviation is the square root of the variance:
\(\displaystyle\sigma=\sqrt{{{\sigma_{{Y}}^{{2}}}+{\sigma_{{X}}^{{2}}}}}=\sqrt{{{0.4}^{{2}}+{0.3}^{{2}}}}={0.5}\)
c) Determine the z-score:
\(\displaystyle{z}={\frac{{{0}-{1}}}{{{0.5}}}}=-{2}\)
Look up the corresponding probability in the table
\(\displaystyle{P}{\left({Y}-{X}{<}{0}\right)}={0.0228}={2.28}\%\)
If X is the random variable representing the small bowl and Y the random variable representing the large bowl, then Y+X represents the total amount of the large and small bowl.
d) The expected value changes the same as the random variables:
\(\displaystyle\mu=\mu_{{Y}}+\mu_{{X}}={2.5}+{1.5}={4}\)
The variance changes by the square of the coefficients and the standard deviation is the square root of the variance:
\(\displaystyle\sigma=\sqrt{{{\sigma_{{Y}}^{{2}}}+{\sigma_{{X}}^{{2}}}}}=\sqrt{{{0.4}^{{2}}+{0.3}^{{2}}}}={0.5}\)
e) Determine the z-score
\(\displaystyle{z}={\frac{{{4.5}-{4}}}{{{0.5}}}}={1}\)
Look up the corresponding probability in the table
\(\displaystyle{P}{\left({Y}+{X}{>}{4.5}\right)}={1}-{0.8413}={0.1587}={15.87}\%\)
f) If X is the random variable representing the small bowl and Y the random variable representing the large bowl and Z represents the amount in the box, then \(\displaystyle{Z}-{\left({X}+{Y}\right)}\) represents the total amount left in the box.
The expected value changes the same as the random variables:
\(\displaystyle\mu=\mu_{{Z}}-\mu_{{X}}-\mu_{{Y}}={16.3}-{1.5}-{2.5}={12.3}\)
The variance changes by the square of the coefficients and the standard deviation is the square root of the variance:
\(\displaystyle\sigma=\sqrt{{{\sigma_{{Z}}^{{2}}}+{\sigma_{{Y}}^{{2}}}+{\sigma_{{X}}^{{2}}}}}=\sqrt{{{0.2}^{{2}}+{0.4}^{{2}}+{0.3}^{{2}}}}\approx{0.54}\)

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