Question

7) In Problem 6 we poured a large and a small bowl of cereal from a box. Suppose the amount of cereal that the manufacturer puts in the boxes is a random variable with mean 16.2 ounces and standard deviation 0.1 ounces. a) Find the expected amount of cereal left in the box. b) What’s the standard deviation? c) If the weight of the remaining cereal can be described by a Normal model, what’s the probability that the box still contains more than 13 ounces?

Answer 1

Question:

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.

Answer:

a) The expected amount of cereal left in the box is 12.2 ounces

b) The standard deviation  $\sigma_{x+y+z}$, is 0.5099

c) In a Normal model, the probability that the box still contains more than 13 ounces is P(Z-(X+Y) > 13) = 5.821 %.

Step-by-step explanation:

Let X represent the amount of cereal that can be poured into a small bowl and Y represent the amount of cereal that can be poured into a large bowl and Z represent the amount of cereal that the manufacturer puts in the box, then the expected amount of cereal left in the box is given by

Z - (X + Y)

(a) The expected amount of cereal left in the box is given as

P(Z - (X + Y)) = μ = μ$_Z$ - μ$_X$ - μ$_Y$ = 16.2 - 1.5 - 2.5 = 12.2 ounces

The expected amount of cereal left in the box = 12.2 ounces

b) The standard deviation is given by  the root of the sum of the variance

That is

$\sigma_{x+y+z} ^2 = \sigma_x^2 + \sigma_y^2 +\sigma_z^2$ and

$\sqrt{\sigma_{x+y+z} ^2} = \sqrt{\sigma_x^2 + \sigma_y^2 +\sigma_z^2} =\sqrt{0.1^2+0.4^2+0.3^2}$ = 0.5099

The standard deviation,  $\sigma_{x+y+z}$, = 0.5099

c) The probability that the box still contains more than 13 ounces is given by

P(Z-(X+Y) > 13)

Where z-score is  

$z=\frac{x-\mu}{\sigma} = \frac{13-12.2}{0.5099}= 1.5689$ ≈ 1.57

From the z-score table P(Z = 1.57) = 0.94179

Therefore the probability of the box containing ≤ 13 is 0.94179, that is

P(Z-(X+Y) ≤ 13) = 0.94179 and

P(Z-(X+Y) > 13) = 1 - 0.94179 = 0.05821 = 5.821 %.