Several friends each had 2/5 of a bag of peanuts left over from the baseball game. they realised that they could have bought 2 f
2 answers:
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Answer:
(a) The expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.
(b) The expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.
(c) Because the variance of each variable is dependent on the other.
Step-by-step explanation:
The random variable <em>X</em> and <em>Y</em> are defined as follows:
<em>X</em> = amount of ice cream in the box
<em>Y</em> = amount of ice cream scooped out
The information provided is:
E (X) = 48
SD (X) = 1
V (X) = 1
E (Y) = 2
SD (Y) = 0.25
V (Y) = 0.0625
(a)
The total amount of ice-cream served at the party can be expressed as:
<em>X</em> + 3<em>Y</em>.
Compute the expected value of (<em>X</em> + 3<em>Y</em>) as follows:
Compute the variance of (<em>X</em> + 3<em>Y</em>) as follows:
Then the standard deviation of (<em>X</em> + 3<em>Y</em>) is:
Thus, the expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.
(b)
The amount of ice-cream left in the box after scooping out one scoop is represented as follows:
<em>X</em> - <em>Y</em>.
Compute the expected value of (<em>X</em> - <em>Y</em>) as follows:
Compute the variance of (<em>X</em> - <em>Y</em>) as follows:
Then the standard deviation of (<em>X</em> - <em>Y</em>) is:
Thus, the expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.
(c)
The variance of the sum or difference of two variables is computed by adding the individual variances. This is because the variance of each variable is dependent on the others.