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Sagot :
Answer:
If the question has
What are the mean and standard deviation of the sampling distribution of the difference in sample means x¯F−x¯M ?
It is E
Step-by-step explanation:
AP classroom uses the example with different end questions.
The mean and standard deviation of the sampling distribution of the difference in sample means xF − xM is 1.5 minutes(mean) and 1.533 minutes(std. dev) approx The probability of getting a difference in sample means xF − xM that is less than 0 is 0.0239 approx.
What is the distribution of random variable which is sum of normal distributions?
Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:
[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]
And if
[tex]X = X_1 + X_2 + \cdots + X_n[/tex]
Then, its distribution is given as:
[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]
For this case, it is given that:
- Distribution of time taken for distance ran by female is Normal distribution and had a mean running time of [tex]\mu_F[/tex] = 8.8 minutes with standard deviation of [tex]\sigma_F[/tex] = 3.3 minutes.
- Distribution of time taken for distance ran by male is Normal distribution and had a mean running time of [tex]\mu_M[/tex] = 7.3 minutes with standard deviation of [tex]\sigma_M[/tex] = 2.9 minutes.
- Samples of size n = 8 are taken from both population.
Estimated sample mean = population mean
Estimated sample standard deviation = population std. dev / √n
where n = sample size
Thus, we get:
- [tex]\overline{x}_F = \mu_F = 8.8 \: \rm minutes[/tex]
- [tex]s_F = \sigma_F/\sqrt{8} = 3.3/\sqrt{8} \approx 1.167 \: \rm minutes[/tex]
- [tex]\overline{x}_M = \mu_M = 7.3 \: \rm minutes[/tex]
- [tex]s_M = \sigma_M/\sqrt{8} = 2.9/\sqrt{8} \approx 1.053\: \rm minutes[/tex]
For sample of females, the distribution is [tex]X_F \sim N(\overline{x}_F = 8.8, s_F = 3.3/\sqrt{8})[/tex]
For sample of males, the distribution is [tex]X_M \sim N(\overline{x}_M =7.3, s_M = 2.9/\sqrt{8})[/tex]
Also, if we take distribution of difference between sample means([tex]X_F - X_M = X_F + (-X_M)[/tex], then it would be normal with mean = 8.8-7.3 = 1.5 minutes (as mean for [tex]-X_F[/tex] will be -7.3 but same standard deviation) as factor gets squared for affecting standard deviation and variance)
and standard deviation is: [tex]s = \sqrt{s_F^2 + s_M^2} = \sqrt{3.3^2/8 + 2.9^2/8} \approx 1.553 \: \rm minutes[/tex]
The distribution of difference between sample means is [tex]X_F - X_M \sim N(\overline{x} = 1.5, s = 1.533)[/tex] approximately
We need [tex]P(X_F - X_M < 0)[/tex]
Converting this distribution to standard normal distribution, we get:
[tex]P(X_F - X_M < 0) = P(Z < \dfrac{0 - \overline{x}}{s} \approx -1.98)[/tex]
From the z-tables, we get the p value for Z = -1.98 as 0.0239
Thus, we get:
[tex]P(X_F - X_M < 0) = P(Z < \dfrac{0 - \overline{x}}{s} \approx -1.98) \approx 0.0239[/tex]
Thus, the mean and standard deviation of the sampling distribution of the difference in sample means xF − xM is 1.5 minutes(mean) and 1.533 minutes(std. dev) approx The probability of getting a difference in sample means xF − xM that is less than 0 is 0.0239 approx.
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
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