Answer:
The warranty period should be more than 4.23 years such that 98% of the players need replacement after the warranty expires.
Explanation:
Let's X = The time after which the CD player needs replacement.
It is provided that,
[tex]X\sim N(\mu = 7.1\ years,\sigma=1.4\ years)[/tex]
Let (X > T) denote the event that the players need the replacement after T years.
It is provided that,
P (X > T) = 0.98.
Determine the value of T as follows:
[tex]P(X>T)=0.98\\P(\frac{X-\mu}{\sigma}>\frac{T-7.1}{1.4})=0.98\\ P(Z>z)=0.98\\1-P(Z\leq z)=0.98\\P(Z\leq z)=0.02[/tex]
Use the standard normal table to compute the value of z.
The value of z is -2.05.
Compute the value of T as follows:
[tex]\frac{T-7.1}{1.4} =-2.05\\T=7.1-(2.05\times1.4)\\=4.23[/tex]
Thus, the warranty period should be more than 4.23 years.
In this exercise we have to use the knowledge of statistics to calculate the time needed for the warranty period, in this way we find that:
The promise period bear happen more than 4.23 old age specific that 98% of the players need substitute after the promise expires.
In this way we can use the following data to set up the equation as:
We can call X as the time after which the CD player needs replacement, so will be like:
[tex]X=N (\mu= 7.1, \sigma=1.4)[/tex]
Let (X > T) mean the occurrence that the players need the substitute following in position or time T old age, will be:
[tex]P (X > T) = 0.98[/tex]
Now performing the calculation to determine T, we find that:
[tex]P(X>T)=0.98\\P(\frac{X-\mu}{\sigma}>\frac{T-7.1}{1.4})= 0.98\\P(Z>z)=0.98\\1-P=0.98\\P=0.02[/tex]
Use the standard usual table to calculate the advantage of z. The profit of z exist -2.05. Compute the profit of T in this manner:
[tex]\frac{T-7.1}{1.4}=-2.05\\T=7.1-(2.05*1.4)= 4.23[/tex]
See more about statistics at brainly.com/question/10951564
