A large school district held a district-wide track meet for all high school students. For the 2-mile run, the population of female students participating follows a Normal distribution and had a mean running time of 8.8 minutes with standard deviation of 3.3 minutes. The population of male students participating also follows a Normal distribution and had a mean running time of 7.3 minutes with standard deviation 2.9 minutes. Suppose 8 female students and 8 male students who participated in the 2-mile run are selected at random from each population. Let xF represent the sample mean running time for the female students, and let xM represent the sample mean running time for the male students. a. Find and interpret the mean and standard deviation of the sampling distribution of the difference in sample means xF − xM. b. Find the probability of getting a difference in sample means xF − xM that is less than 0.

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.

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:

Estimated sample mean  = population mean

Estimated sample standard deviation = population std. dev / √n

where n = sample size

Thus, we get:

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:

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