Respuesta :
Answer:
a) 25.78% probability that a sample of 30 of the drivers will have a sample mean less than 56 MPH
b) 21.48% probability that a sample of 45 of the drivers will have a sample mean less than 56 MPH
c) 18.14% probability that a sample of 60 of the drivers will have a sample mean less than 56 MPH
d)
i. As the sample size increases, the standard error of the mean
decreases
ii. As the sample size increases, the standard error of the mean
move farther the population mean of 57.3 MPH.
iii. Therefore, the probability of observing a sample mean less than 56 MPH
decrease.
Step-by-step explanation:
To solve this question, we have 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 56.6, \sigma = 5.1[/tex]
a. What is the probability that a sample of 30 of the drivers will have a sample mean less than 56 MPH?
This means that [tex]s = \frac{5.1}{\sqrt{30}} = 0.93[/tex]
This probability is the pvalue of Z when X = 56. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 56.6}{0.93}[/tex]
[tex]Z = -0.65[/tex]
[tex]Z = -0.65[/tex] has a pvalue of 0.2578.
25.78% probability that a sample of 30 of the drivers will have a sample mean less than 56 MPH
b. What is the probability that a sample of 45 of the drivers will have a sample mean less than 56 MPH?
[tex]s = \frac{5.1}{\sqrt{45}} = 0.76[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 56.6}{0.76}[/tex]
[tex]Z = -0.79[/tex]
[tex]Z = -0.79[/tex] has a pvalue of 0.2148.
21.48% probability that a sample of 45 of the drivers will have a sample mean less than 56 MPH
c. What is the probability that a sample of 60 of the drivers will have a sample mean less than 56 MPH?
[tex]s = \frac{5.1}{\sqrt{60}} = 0.66[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 56.6}{0.66}[/tex]
[tex]Z = -0.91[/tex]
[tex]Z = -0.91[/tex] has a pvalue of 0.1814.
18.14% probability that a sample of 60 of the drivers will have a sample mean less than 56 MPH
d. Explain the difference in these probabilities. (Select one each)
n increases, so s decreases(moving farther from the mean), which means that z decreases, leaving 56 mph to a lower percentile. So
i. As the sample size increases, the standard error of the mean
decreases
ii. As the sample size increases, the standard error of the mean
move farther the population mean of 57.3 MPH.
iii. Therefore, the probability of observing a sample mean less than 56 MPH
decrease.
