Respuesta :
Answer:
0.2061 = 20.61% probability that the mean of the women will be more than 1.5 points higher than the mean of the men
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 normal distributions can be 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.
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Women:
Mean 5, standard deviation 1.5.
Sample of 12.
So
[tex]\mu_{W} = 5[/tex]
[tex]s_{W} = \frac{1.5}{\sqrt{12}} = 0.433[/tex]
Men:
Mean of 4, standard deviation 1.5.
Sample of 12.
So
[tex]\mu_{M} = 4[/tex]
[tex]s_{M} = \frac{1.5}{\sqrt{12}} = 0.433[/tex]
What is the probability that the mean of the women will be more than 1.5 points higher than the mean of the men?
This is:
[tex]P(W - M) \geq 1.5[/tex]
Subtraction:
[tex]\mu_{W-M} = \mu_{W} - \mu_{M} = 5 - 4 = 1[/tex]
[tex]s_{W - M} = \sqrt{s_{W}^2+S_{M}^2} = \sqrt{0.433^2+0.433^2} = 0.6124[/tex]
The probability is 1 subtracted by the pvalue of Z when X = 1.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this distribution
[tex]Z = \frac{X - \mu_{W-M}}{s_{W - M}}[/tex]
[tex]Z = \frac{1.5 - 1}{0.6124}[/tex]
[tex]Z = 0.82[/tex]
[tex]Z = 0.82[/tex] has a pvalue of 0.7939
1 - 0.7939 = 0.2061
0.2061 = 20.61% probability that the mean of the women will be more than 1.5 points higher than the mean of the men
