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8. Daren and Josh are pretty good free throw shooters. Daren makes 75% of the
free throws he attempts. Josh makes 80% of his free throws. Suppose we take
separate random samples of 50 free throws each from Daren and Josh, and
record the proportion of free throws that are made by each. Which of the
following best describes the sampling distribution of PD - Pj?

Answer :

Answer:

B) Approximately normal, with mean -0.05 and standard deviation 0.083

Step-by-step explanation:

The correct answer is (B). The shape is approximately normal since the expected number of makes and misses for both Daren and Josh are all greater than 10.

The distribution of pD = pJ is given by: Option B: Approximately normal, with mean -0.05 and standard deviation 0.083

What is the distribution of population proportion?

For large enough sample, let the population proportion of a quantity be denoted by random variable [tex]p[/tex]

Then, we get:

[tex]p \sim N(\hat{p}, \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})[/tex]

where

  • [tex]\hat{p}[/tex] = estimated(mean value) proportion of that quantity, and
  • n = size of sample drawn.


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]

Here, we're specified that:

For proportion of correct free throw attempts by Daren:

  • Sample size = n = 50
  • Estimated proportion = [tex]\hat{p}_D = 75\% = 0.75[/tex]
  • Let the population proportion be represented by variable [tex]p_D[/tex]

For proportion of correct free throw attempts by Josh:

  • Sample size = n = 50
  • Estimated proportion = [tex]\hat{p}_J = 80\% = 0.8[/tex]
  • Let the population proportion be represented by variable [tex]p_J[/tex]

And therefore, we have:

[tex]p_D \sim N\left (0.75, \sqrt{\dfrac{0.75 \times 0.25}{50}}\right )\\\\p_J \sim N\left (0.80, \sqrt{\dfrac{0.8 \times 0.2}{50}}\right )[/tex]

And therefore, we get:

[tex]-p_J \sim N\left (-0.80, \sqrt{\dfrac{0.8 \times 0.2}{50}}\right )[/tex] (standard deviation doesn't get affected by change of sign of the considered random variable, and mean gets affected linearly, so mean became negative 0.80 and standard deviation stays same).

Adding [tex]p_D[/tex] and [tex]-p_J[/tex], we get:

[tex]p_D + (-p_J) = p_D - p_j \sim N\left (0.75 -0.80, \sqrt{\dfrac{0.8 \times 0.2}{50} + \dfrac{0.75 \times 0.25}{50}}\right )\\\\p_D - p_J \sim N\left( -0.05, 0.0837\right)[/tex]

Thus, the distribution of pD = pJ is given by: Option B: Approximately normal, with mean -0.05 and standard deviation 0.083

Learn more about population proportion here:

https://brainly.com/question/7204089

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