Respuesta :
Answer:
(A) The shape of the distribution of the sample mean is bell-shaped.
(B) The standard error of the distribution of the sample mean is 1.1.
(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.
(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.
(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.
Step-by-step explanation:
We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.
As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
Let [tex]\bar X[/tex] = sample mean life of these batteries
(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.
(B) The standard error of the distribution of the sample mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
So, the standard error = [tex]\frac{5.5}{\sqrt{25} }[/tex] = 1.1.
(C) The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean life of battery = 35.0 hours
[tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)
P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)
= 1 - 0.8186 = 0.1814
(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)
P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)
= 0.6736
(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)
P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)
P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186
P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)
= 1 - 0.6736 = 0.3264
Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.
