Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
0% probability that the mean score for the B group will be at least five points higher than the mean score for the A group
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution, the central limit theorem, and subtraction of normal variables:
Normal probability distribution
When the distribution is normal, we use 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.
Subtraction of normal variables:
When we subtract two normal variables, the mean is the subtraction of the mean of each variable, while the standard deviation is the square root of the sum of each variance.
Method A:
One student: N(75,30), so [tex]\mu = 75, \sigma = 30[/tex]
Samples of 100 students: [tex]n = 100, s = \frac{30}{\sqrt{100}} = 0.3[/tex]
Method B:
One student: N(74,40), so [tex]\mu = 74, \sigma = 40[/tex]
Samples of 100 students: [tex]n = 100, s = \frac{40}{\sqrt{100}} = 0.4[/tex]
What is the probability that the mean score for the B group will be at least five points higher than the mean score for the A group?
This is the probability that B - A >= 5. So
[tex]\mu_{B-A} = \mu_B - \mu_A = 74 - 75 = -1[/tex]
[tex]s_{B-A} = \sqrt{s_A^{2}+S_B^{2}} = \sqrt{(0.3)^2+(0.4)^2} = 0.5[/tex]
We have to find 1 subtracted by the pvalue of Z when X = 5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5 - (-1)}{0.5}[/tex]
[tex]Z = 12[/tex]
[tex]Z = 12[/tex] has a pvalue of 1
1 - 1 = 0, so
0% probability that the mean score for the B group will be at least five points higher than the mean score for the A group
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
