College

A nationwide test taken by high school sophomores and juniors has three sections, each scored on a scale of 20 to 80. In a recent year, the national mean score for the writing section was 47.1, with a standard deviation of 9.9.

Based on this information, complete the following statements about the distribution of the scores on the writing section for the recent year:

1. According to Chebyshev's theorem, at least what percent of the scores lie between 27.3 and 66.9?
2. According to Chebyshev's theorem, at least 84% of the scores lie between what two values?

(Round your answer to 1 decimal place.)

Answer :

Answer:

Step-by-step explanation:

Problem 1: Chebyshev's Theorem and Scores Between 27.3 and 66.9

Given:

Mean (μ) = 47.1

Standard Deviation (σ) = 9.9

Interval: [27.3, 66.9]

1. Calculate the distance from the mean to the interval boundaries in terms of standard deviations:

k = ∣27.3−47.1∣ / 9.9 = 19.8 / 9.9 =2

Similarly:

k = ∣66.9−47.1∣ / 9.9 =2

So, the interval [27.3, 66.9] corresponds to k=2 standard deviations.

2. Apply Chebyshev's theorem: According to Chebyshev's theorem, at least

1 − 1/k^2 of the data lies within k standard deviations of the mean:

1 − 1/2^2 = 1 − 1/4 = 0.75

Multiply by 100 to get the percentage:

0.75×100=75%

Answer: At least 75% of the scores lie between 27.3 and 66.9.

Problem 2: Interval for At Least 84% of the Scores

Given:

Percent of data: 84%

Minimum percentage covered by Chebyshev's theorem:

1 − 1/^2

1. ​Solve for k:

1 − 1/^2 = 0.84

1/^2 = 0.16

k ^2 = 1/0.16 = 6.25

k = (6.25)^1/2 = 2.5

2. Calculate the interval: The range is μ±kσ, so:

Lower bound=μ−2.5σ=47.1−2.5(9.9)=47.1−24.75=22.4

Upper bound=μ+2.5σ=47.1+2.5(9.9)=47.1+24.75=71.9

Answer: At least 84% of the scores lie between 22.4 and 71.9.

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