A civil service exam yields scores with a mean of 81 and a standard deviation of 5.5. Using Chebyshev's Theorem what can we say about the percentage of scores that are above 92?

At most 12.5% of the scores are above 92.

Step-by-step explanation:

Chebyshev's theorem states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 81%

Standard deviation = 5.5%

Using Chebyshev's Theorem what can we say about the percentage of scores that are above 92?

92 = 81 + 2*5.5

So 92 is two standard deviations above the mean.

By the Chebyshev's theorem, at least 75% of the measures are within 2 standard deviations of the mean. So at most 25% is more than 2 standard deviations from the mean. Chebyshev's theorem works with symmetric distributions, so, of those at most 25%, at most 12.5% are more than 2 standard deviations below the mean and at most 12.5% are more than 2 standard deviations above the mean.

So at most 12.5% of the scores are above 92.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°