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?

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.

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

b) The standard deviation  , 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 - μ - μ = 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

and

= 0.5099

The standard deviation,  , = 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  

≈ 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 %.

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

step-by-step explanation: