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.

Answer 1

Answer:

0.119 is the probability that their scores differ by more than 15 points.

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 41

Standard Deviation, σ = 9

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

D is a random variable and defined as the difference between their scores.

$D = X-Y\\\mu_D = E(X)-E(Y) = 41-41 = 0\\ \sigma_D = \sqrt{\sigma_x^2 + \sigma_y^2 } = \sqrt{9^2 + 9^2} = 9\sqrt{2}$

D follows a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(scores differ by more than 15 points)

$P(d > 15) = P(z > \displaystyle\frac{15-0}{9\sqrt{2}}) = P(z > 1.1785)\\\\P( z > 1.1785) = 1 - P(z \leq 1.1785)$

Calculating the value from the standard normal table we have,

$1 - 0.881 = 0.119 = 11.9\%\\P( d > 15) = 11.9\%$

0.119 is the probability that their scores differ by more than 15 points.