Respuesta :
Answer:
a) [tex]\bar X \sim N(\mu=48, \frac{\sigma}{\sqrt{n}}=\frac{8.2}{\sqrt{8}}=2.899)[/tex]
b) [tex]P(\bar X <42.2)=P(Z<-2.00)=0.0228[/tex]
Step-by-step explanation:
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Part a
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]\bar X \sim N(\mu=48, \frac{\sigma}{\sqrt{n}}=\frac{8.2}{\sqrt{8}}=2.899)[/tex]
Part b
We want this probability
[tex]P(\bar X <42.2)[/tex]
In order to answer this question we can use the z score in order to find the probabilities, the formula given by:
[tex]z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\bar X <42.2)=P(Z<\frac{42.2-48}{\frac{8.2}{\sqrt{8}}}=-2.00)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-2.00)=0.0228[/tex]
(a) The sample mean lifetime is 2.899 months
(b) The probability that the sample mean lifetime is 42.2 months or less if the lifetime distribution is unchanged is 0. 0228.
The normal distribution for a random variable X is defined as
[tex]\rm f(x) = \dfrac{1}{\sigma \sqrt{2\pi}} (e^{(-1/2) (\dfrac{x-\mu}{\sigma} )^2 })}\\where\\\sigma = Population \; standard \; deviation\\\mu = Population\; mean[/tex]
According to the given situation
[tex]\rm \mu = 48\; month\\\sigma = 8.2 \; month \\[/tex]
(a) If the new brand has the same lifetime distribution as the previous type of battery,
The sampling distribution of the mean lifetime can be expressed as
[tex]\rm \bar X = \dfrac{\sigma }{\sqrt{n}}[/tex]
So we can write
[tex]\rm \bar X = \dfrac{8.2}{\sqrt{8}} = 8.2/2.828 = 2.899[/tex]
So the sample mean lifetime is 2.899 months
(b) The average life of the batteries on these 8 cars turns out to be = 42.2 months.
We have to find out the probability that the sample mean lifetime is 42.2 months or less if the lifetime distribution is unchanged.
So according to the language of question we have to determine
[tex]\rm P(\bar X<42.2)[/tex]
also we know that
[tex]\rm z = (x -\mu)/\sigma[/tex]
putting the values of
[tex]\rm \bar X ,\mu , \sigma \; we \; can \; find\; z = (42.2 -48)/2.89[/tex]
z = -2.00
[tex]\rm P(z<- 2.00) = 0.0228[/tex]
Conclusions are as follows
(a) The sample mean lifetime is 2.899 months
(b) The probability that the sample mean lifetime is 42.2 months or less if the lifetime distribution is unchanged is 0. 0228.
For more information please refer to the link below
https://brainly.com/question/25394084
