Some nations require their students to pass an exam before earning their primary school degrees or diplomas. A certain nation gives students an exam whose scores are normally distributed with a mean of 41 points and a standard deviation of 9 points.

Suppose we select 2 of these testers at random, and define the random variable $D$ as the difference between their scores. We can assume that their scores are independent.

Find the probability that their scores differ by more than 15 points. You may round your answer to two decimal places.
$$
P(|D|15) \approx
$$
$\square$

Q: Some nations require their students to pass an exam before earning their primary school degrees or diplomas. A certain nation gives students an exam whose scores are normally distributed with a mean of 41 points and a standard deviation of 9 points. Suppose we select 2 of these testers at random, and define the random variable $D$ as the difference between their scores. We can assume that their scores are independent. Find the probability that their scores differ by more than 15 points. You may round your answer to two decimal places. \[ P(|D|>15) \approx \] $\square$

∻Solution by Steps∻ ‖ step 1 ⋮ Define the random variable $ D $ as the difference between the scores of two randomly selected testers. Since the scores are normally distributed with mean $ \mu = 41 $ and standard deviation $ \sigma = 9 $, the difference $ D $ will also be normally distributed with mean $ \mu_D = 0 $ and standard deviation $ \sigma_D = \sqrt{2} \cdot \sigma = \sqrt{2} \cdot 9 = 9\sqrt{2} $. ‖ ‖ step 2 ⋮ We need to find the probability that the absolute value of $ D $ is greater than 15, i.e., $ P(|D| > 15) $. This can be written as $ P(D > 15) + P(D 1.18) $. This is approximately $ 1 - 0.8810 = 0.1190 $. ‖ ‖ step 5 ⋮ Since the normal distribution is symmetric, $ P(Z 1.18) $. Therefore, $ P(|D| > 15) = 2 \cdot P(Z > 1.18) = 2 \cdot 0.1190 = 0.2380 $. ‖ ∻Answer∻ ⚹ 0.24 ⚹ ∻Key Concept∻ ⚹The difference between two normally distributed independent variables is also normally distributed.⚹ ∻Explanation∻ ⚹We used the properties of the normal distribution and standardization to find the probability that the absolute difference between two scores exceeds 15 points.⚹◊What is the formula for calculating the probability of the absolute difference between two normally distributed random variables?⍭ Generate me a similar question◊

Question

Some nations require their students to pass an exam before earning their primary school degrees or diplomas. A certain nation gives students an exam whose scores are normally distributed with a mean of 41 points and a standard deviation of 9 points.

Suppose we select 2 of these testers at random, and define the random variable $D$ as the difference between their scores. We can assume that their scores are independent.

Find the probability that their scores differ by more than 15 points. You may round your answer to two decimal places.
$$
P(|D|>15) \approx
$$
$\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Define the random variable $ D $ as the difference between the scores of two randomly selected testers. Since the scores are normally distributed with mean $ \mu = 41 $ and standard deviation $ \sigma = 9 $, the difference $ D $ will also be normally distributed with mean $ \mu_D = 0 $ and standard deviation $ \sigma_D = \sqrt{2} \cdot \sigma = \sqrt{2} \cdot 9 = 9\sqrt{2} $. ‖ ‖ step 2 ⋮ We need to find the probability that the absolute value of $ D $ is greater than 15, i.e., $ P(|D| > 15) $. This can be written as $ P(D > 15) + P(D < -15) $. ‖ ‖ step 3 ⋮ Standardize the variable $ D $ to the standard normal distribution $ Z $ using the formula $ Z = \frac{D - \mu_D}{\sigma_D} $. For $ D = 15 $, $ Z = \frac{15 - 0}{9\sqrt{2}} = \frac{15}{9\sqrt{2}} = \frac{15}{9 \cdot 1.414} \approx 1.18 $. ‖ ‖ step 4 ⋮ Using the standard normal distribution table, find $ P(Z > 1.18) $. This is approximately $ 1 - 0.8810 = 0.1190 $. ‖ ‖ step 5 ⋮ Since the normal distribution is symmetric, $ P(Z < -1.18) = P(Z > 1.18) $. Therefore, $ P(|D| > 15) = 2 \cdot P(Z > 1.18) = 2 \cdot 0.1190 = 0.2380 $. ‖ ∻Answer∻ ⚹ 0.24 ⚹ ∻Key Concept∻ ⚹The difference between two normally distributed independent variables is also normally distributed.⚹ ∻Explanation∻ ⚹We used the properties of the normal distribution and standardization to find the probability that the absolute difference between two scores exceeds 15 points.⚹◊What is the formula for calculating the probability of the absolute difference between two normally distributed random variables?⍭ Generate me a similar question◊

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