Chapter 12.6, Problem 3CYU

a.

To determine

To calculate:The probability that a randomly chosen customer bought 5 or 6 components

Answer to Problem 3CYU

Theprobability that a randomly chosen customer bought 5 or 6 components is $20\%$

Explanation of Solution

Given information:

Number of customers bought 0 to 2 components are 26

Number of customers bought 3 to 4 components are 42

Number of customers bought5 to 6 components are 33

Number of customers bought 7 to 8 components are 24

Number of customers bought 9 to 10 components are 40

Calculation:

The probability that a randomly chosen customer bought 5 or 6 components can be calculated as,

  $\begin{array}{l}\left(\frac{33}{26+42+33+24+40}\right)\times 100\%\\ =\left(\frac{33}{165}\right)\times 100\%\\ =\left(0.2\times 100\right)\%\\ =20\%\end{array}$

Therefore, the probability that a randomly chosen customer bought 5 or 6 components is $20\%$

b.

To determine

To calculate: The probability that a randomly chosen customer bought less than 5 components

Answer to Problem 3CYU

The probability for customer who have bought less than 5 components is $41.21\%$

Explanation of Solution

Given information:

Number of customers bought 0 to 2 components are 26

Number of customers bought 3 to 4 components are 42

Number of customers bought 5 to 6 components are 33

Number of customers bought 7 to 8 components are 24

Number of customers bought 9 to 10 components are 40

Calculation: the probability for customers who have bought less than 5 components can be calculated as,

  $\begin{array}{l}\left(\frac{26+42}{26+42+33+24+40}\right)\times 100\%\\ =\left(\frac{68}{165}\right)\times 100\%\\ =0.4121\times 100\%\\ =41.21\%\end{array}$

Therefore, the probability for customer who have bought less than 5 components is $41.21\%$