The manager of a music store has kept records of
The manager of a music store has kept records of the number of CDs bought in a single transaction by customers who make a purchase at the store. The accompanying table gives six possible outcomes and the estimated probability associated with each of these outcomes for the chance experiment that consists of observing the number of CDs purchased by a randomly selected customer at the store.
Number of CDs purchased 1 2 3 4 5 6 or more
Estimated probability 0.30 0.25 0.20 0.15 0.05 0.05
(a) What is the estimated probability that the next customer purchases three or fewer CDs?
(b) What is the estimated probability that the next customer purchases at most three CDs?
How does this compare to the probability computed in part (a)?
greater than part (a)less than part (a) the same as part (a)
(c) What is the estimated probability that the next customer purchases five or more CDs?
(d) What is the estimated probability that the next customer purchases one or two CDs?
(e) What is the estimated probability that the next customer purchases more than two CDs?

Answer:

(a) [tex]P(x\le 3) = 0.75[/tex]

(b) [tex]P(x\le 3) = 0.75[/tex]

(b) is the same as (a)

(c) [tex]P(x \ge 5) = 0.10[/tex]

(d) [tex]P(x = 1\ or\ 2) = 0.55[/tex]

(e) [tex]P(x > 2) = 0.45[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}[/tex]

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

[tex]\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}[/tex]

This probability is calculated as:

[tex]P(x\le 3) = P(1) + P(2) + P(3)[/tex]

This gives:

[tex]P(x\le 3) = 0.30 + 0.25 + 0.20[/tex]

[tex]P(x\le 3) = 0.75[/tex]

Solving (b): Probability of at most 3 CDs

Here, we consider:

[tex]\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}[/tex]

This probability is calculated as:

[tex]P(x\le 3) = P(1) + P(2) + P(3)[/tex]

This gives:

[tex]P(x\le 3) = 0.30 + 0.25 + 0.20[/tex]

[tex]P(x\le 3) = 0.75[/tex]

(b) is the same as (a)

Solving (c): Probability of 5 or more CDs

Here, we consider:

[tex]\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}[/tex]

This probability is calculated as:

[tex]P(x \ge 5) = P(5) + P(6\ or\ more)[/tex]

This gives:

[tex]P(x\ge 5) = 0.05 + 0.05[/tex]

[tex]P(x \ge 5) = 0.10[/tex]

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

[tex]\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}[/tex]

This probability is calculated as:

[tex]P(x = 1\ or\ 2) = P(1) + P(2)[/tex]

This gives:

[tex]P(x = 1\ or\ 2) = 0.30 + 0.25[/tex]

[tex]P(x = 1\ or\ 2) = 0.55[/tex]

Solving (e): Probability of more than 2 CDs

Here, we consider:

[tex]\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}[/tex]

This probability is calculated as:

[tex]P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)[/tex]

This gives:

[tex]P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05[/tex]

[tex]P(x > 2) = 0.45[/tex]

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