Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.7 Junior Find the answer

Given a random variable $X \sim N(\mu, \sigma^2)$ ( $\sigma > 0$ ), the following statements are true:
- $P(\mu-\sigma < X \leq \mu+\sigma) = 0.6826$ ,
- $P(\mu-2\sigma < X \leq \mu+2\sigma) = 0.9544$
- $P(\mu-3\sigma < X \leq \mu+3\sigma) = 0.9974$ .

In a class of 60 students, the scores of a math test follow a normal distribution with a mean of 110 and a variance of 100. Theoretically, the number of students scoring between 120 and 130 is approximately:

Pick one

Solution

Since the math scores approximately follow a normal distribution $N(110, 10^2)$ ,
- $P(100 < x < 120) = 0.6826$ , $P(90 < x < 130) = 0.9544$ .

Based on the symmetry of the normal curve, the probability of scoring between 120 and 130 is $\frac{1}{2}(0.9544 - 0.6826) = 0.1359$ .

Therefore, theoretically, the number of students scoring between 120 and 130 is $0.1359 \times 60 \approx 8$ .

Hence, the correct choice is $\boxed{C}$ .

The values of a normal distribution are symmetric about $x=110$ . Using $P(100 < x < 120) = 0.6826$ and $P(90 < x < 130) = 0.9544$ , we can obtain the required result. A random variable that is the sum of many independent, insignificant, and random factors tends to follow or closely approximate a normal distribution. The normal distribution holds a significant position in probability and statistics and satisfies the $3\sigma$ rule.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.