Question

A company that manufactures smartphones developed a new battery that has a longer life span than that of atraditional battery. From the date of purchase of a smartphone, the distribution of the life span of the new batteryis approximately normal with mean 30 months and standard deviation 8 months. For the price of $50, thecompany offers a two-year warranty on the new battery for customers who purchase a smartphone. The warrantyguarantees that the smartphone will be replaced at no cost to the customer if the battery no longer works within24 months from the date of purchase.(a) In how many months from the date of purchase is it expected that 25 percent of the batteries will no longerwork? Justify your answer.(b) Suppose one customer who purchases the warranty is selected at random. What is the probability that thecustomer selected will require a replacement within 24 months from the date of purchase because the batteryno longer works?(c) The company has a gain of $50 for each customer who purchases a warranty but does not requirea replacement. The company has a loss (negative gain) of $150 for each customer who purchases a warrantyand does require a replacement. What is the expected value of the gain for the company for each warrantypurchased?

Answer 1

Answer:

1) at 24.64 months 25% of the cellphones battery will fail

2) there is a 22.66% chance it will use the warranty

3) the expected return is 16.01 dollar of gain per warranty

Explanation:

at 25% we have to look into the normal distirbution table for the Z value that acumulates that

we find out is -0.67

now we solve for X value:

$\frac{X - 30}{8} = -0.67\\\frac{X - \mu}{\sigma} = P_z$

X = 24.64

we solve for the provability when X = 24

we convert into a z value and look for the value inthe table

(24 - 30) / 8 = -0.75

probability = 0.226627352 = 22.66%

expected return

$\left[\begin{array}{cccc}State&Return&Probability&Weight\\not use&50&0.7734&38.67\\use&-100&0.2266&-22.66\\&&&\\Total&&1&16.01\\\end{array}\right]$