Respuesta :
Answer:
3% of the time its life will be less than 3.7 years.
Step-by-step explanation:
We are given that a TV manufacturer knows that their TV's have a normally distributed lifespan, with a mean of 5.6 years, and standard deviation of 1 years.
Let X = Lifespan of TV's
So, X ~ Normal([tex]\mu=5.6,\sigma^{2}=1^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X-\mu}{{\sigma} }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean lifespan = 5.6 years
[tex]\sigma[/tex] = standard deviation = 1 year
Now, if a TV is picked at random, we have to find that 3% of the time its life will be less than how many years, that means;
P(X < x) = 0.03 {where x is the required years}
P( [tex]\frac{ X-\mu}{{\sigma} }[/tex] < [tex]\frac{ x-5.6}{{1} }[/tex] ) = 0.03
P(Z < [tex]\frac{ x-5.6}{{1} }[/tex] ) = 0.03
Now, in the z table the critical value of x which represents the below 3% probability area is given as -1.881, i.e;
[tex]\frac{ x-5.6}{{1} }[/tex] = -1.881
x = 5.6 - 1.881
x = 3.7
Hence, 3% of the time its life will be less than 3.7 years.
