Respuesta :
Answer:
60.85% probability that the sample mean is between $1.4 million and $1.5 million per year
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
In millions of dollars.
[tex]\mu = 1.47, \sigma = 0.3, n = 30, s = \frac{0.3}{\sqrt{30}} = 0.0548[/tex]
What is the probability that the sample mean is between $1.4 million and $1.5 million per year?
This is the pvalue of Z when X = 1.5 subtracted by the pvalue of Z when X = 1.4. So
X = 1.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.5 - 1.47}{0.0548}[/tex]
[tex]Z = 0.55[/tex]
[tex]Z = 0.55[/tex] has a pvalue of 0.7088
X = 1.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.4 - 1.47}{0.0548}[/tex]
[tex]Z = -1.28[/tex]
[tex]Z = -1.28[/tex] has a pvalue of 0.1003
0.7088 - 0.1003 = 0.6085
60.85% probability that the sample mean is between $1.4 million and $1.5 million per year
Answer:
[tex] z= \frac{1.4-1.47}{\frac{0.3}{\sqrt{30}}}= -1.278[/tex]
[tex] z= \frac{1.5-1.47}{\frac{0.3}{\sqrt{30}}}= 0.548[/tex]
And we can find the probability with this difference:
[tex] P(-1.278<z<0.548) = P(z<0.548) -P(z<-1.278) =0.708-0.101= 0.607[/tex]
So then the probability that the sample mean is between $1.4 million and $1.5 million per year is 0.607
Step-by-step explanation:
For this case we have the following info given:
[tex] \mu = 1.47[/tex] the true mean for the problem
n =30 represent the sample size
[tex] \sigma = 0.3 millions[/tex] represent the population deviation
And we want to find this probability
[tex] P(1.4< \bar X <1.5)[/tex]
And we can use the z score given by:
[tex] z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z scores for the limits we got:
[tex] z= \frac{1.4-1.47}{\frac{0.3}{\sqrt{30}}}= -1.278[/tex]
[tex] z= \frac{1.5-1.47}{\frac{0.3}{\sqrt{30}}}= 0.548[/tex]
And we can find the probability with this difference:
[tex] P(-1.278<z<0.548) = P(z<0.548) -P(z<-1.278) =0.708-0.101= 0.607[/tex]
So then the probability that the sample mean is between $1.4 million and $1.5 million per year is 0.607
